case study questions on linear programming class 12

CBSE Case Study Questions for Class 12 Maths Linear Programming Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 12 Maths Linear Programming  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 12 Maths Linear Programming PDF

Checkout our case study questions for other chapters.

• Chapter 9 Differential Equations Case Study Questions
• Chapter 10 Vector Algebra Case Study Questions
• Chapter 11 Three dimensional Geometry Case Study Questions
• Chapter 13 Probability Case Study Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

Contact Form

Privacy Policy

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 12 Maths Chapter 12 Linear Programming

• Last modified on: 1 year ago
• Reading Time: 6 Minutes

[PDF] Download Case Study Questions for Class 12 Maths Chapter 12 Linear Programming

Here we are providing case study questions for class 12 maths. In this article, we are sharing Class 12 Maths Chapter 12 Linear Programming case study questions. All case study questions of class 12 maths are solved so that students can check their solutions after attempting questions.

Case Study Questions for Class 12 Maths

Case study questions are a type of question that is commonly used in academic and professional settings to evaluate a person’s ability to analyze, interpret, and solve problems based on a given scenario or case study.

Typically, a case study question presents a real-world situation or problem that requires the individual to apply their knowledge and skills to identify the issues, consider various solutions, and recommend a course of action.

Case study questions are designed to test critical thinking skills, problem-solving abilities, and the capacity to work through complex and ambiguous situations.

Preparing for case study questions involves developing a deep understanding of the subject matter, being able to analyze and synthesize information quickly, and being able to communicate ideas clearly and effectively.

Importance of Solving Case Study Questions for Class 12 Maths

Solving case study questions for Class 12 Maths is extremely important as it provides students with an opportunity to apply the mathematical concepts they have learned to real-world scenarios. These questions present a situation or problem that requires students to use their problem-solving skills and critical thinking abilities to arrive at a solution.

The importance of solving case study questions for Class 12 Maths can be summarized as follows:

• Enhances problem-solving skills: Case study questions challenge students to think beyond textbook examples and apply their knowledge to real-world situations. This enhances their problem-solving skills and helps them develop a deeper understanding of the mathematical concepts.
• Improves critical thinking abilities: Case study questions require students to analyze and evaluate information, and draw conclusions based on their understanding of the situation. This helps them develop their critical thinking abilities, which are essential for success in many areas of life.
• Helps in retaining concepts: Solving case study questions helps students retain the concepts they have learned for a longer period of time. This is because they are more likely to remember the concepts when they have applied them to a real-world situation.
• Better preparation for exams: Many competitive exams, including the Class 12 Maths board exam, contain case study questions. Solving these questions helps students become familiar with the format of the questions and the skills required to solve them, which can improve their performance in exams.

In conclusion, solving case study questions for Class 12 Maths is important as it helps students develop problem-solving and critical thinking skills, retain concepts better, and prepare for exams.

Related Posts

Category lists (all posts).

All categories of this website are listed below with number of posts in each category for better navigation. Visitors can click on a particular category to see all posts related to that category.

• Full Form (1)
• Biography of Scientists (1)
• Assertion Reason Questions in Biology (37)
• Case Study Questions for Class 12 Biology (14)
• DPP Biology for NEET (12)
• Blog Posts (35)
• Career Guidance (1)
• Assertion Reason Questions for Class 10 Maths (14)
• Case Study Questions for Class 10 Maths (15)
• Extra Questions for Class 10 Maths (12)
• Maths Formulas for Class 10 (1)
• MCQ Questions for Class 10 Maths (15)
• NCERT Solutions for Class 10 Maths (4)
• Quick Revision Notes for Class 10 Maths (14)
• Assertion Reason Questions for Class 10 Science (16)
• Case Study Questions for Class 10 Science (14)
• Evergreen Science Book Solutions for Class 10 (17)
• Extra Questions for Class 10 Science (23)
• HOTS for Class 10 Science (17)
• Important Questions for Class 10 Science (10)
• Lakhmir Singh Class 10 Biology Solutions (4)
• Lakhmir Singh Class 10 Chemistry Solutions (5)
• Lakhmir Singh Class 10 Physics Solutions (5)
• MCQ Questions for Class 10 Science (20)
• NCERT Exemplar Solutions for Class 10 Science (16)
• NCERT Solutions for Class 10 Science (15)
• Quick Revision Notes for Class 10 Science (4)
• Study Notes for Class 10 Science (17)
• Assertion Reason Questions for Class 10 Social Science (14)
• Case Study Questions for Class 10 Social Science (24)
• MCQ Questions for Class 10 Social Science (3)
• Topicwise Notes for Class 10 Social Science (4)
• CBSE CLASS 11 (1)
• Assertion Reason Questions for Class 11 Chemistry (14)
• Case Study Questions for Class 11 Chemistry (11)
• Free Assignments for Class 11 Chemistry (1)
• MCQ Questions for Class 11 Chemistry (8)
• Very Short Answer Questions for Class 11 Chemistry (7)
• Assertion Reason Questions for Class 11 Entrepreneurship (8)
• Important Questions for CBSE Class 11 Entrepreneurship (1)
• Assertion Reason Questions for Class 11 Geography (24)
• Case Study Questions for Class 11 Geography (24)
• Assertion Reason Questions for Class 11 History (12)
• Case Study Questions for Class 11 History (12)
• Assertion and Reason Questions for Class 11 Maths (16)
• Case Study Questions for Class 11 Maths (16)
• Formulas for Class 11 Maths (6)
• MCQ Questions for Class 11 Maths (17)
• NCERT Solutions for Class 11 Maths (8)
• Case Study Questions for Class 11 Physical Education (11)
• Assertion Reason Questions for Class 11 Physics (15)
• Case Study Questions for Class 11 Physics (12)
• Class 11 Physics Study Notes (5)
• Concept Based Notes for Class 11 Physics (2)
• Conceptual Questions for Class 11 Physics (10)
• Derivations for Class 11 Physics (3)
• Extra Questions for Class 11 Physics (13)
• MCQ Questions for Class 11 Physics (16)
• NCERT Solutions for Class 11 Physics (16)
• Numerical Problems for Class 11 Physics (4)
• Physics Formulas for Class 11 (7)
• Revision Notes for Class 11 Physics (11)
• Very Short Answer Questions for Class 11 Physics (11)
• Assertion Reason Questions for Class 11 Political Science (20)
• Case Study Questions for Class 11 Political Science (20)
• CBSE CLASS 12 (8)
• Extra Questions for Class 12 Biology (14)
• MCQ Questions for Class 12 Biology (13)
• Case Studies for CBSE Class 12 Business Studies (13)
• MCQ Questions for Class 12 Business Studies (1)
• Revision Notes for Class 12 Business Studies (10)
• Assertion Reason Questions for Class 12 Chemistry (15)
• Case Study Based Questions for Class 12 Chemistry (14)
• Extra Questions for Class 12 Chemistry (5)
• Important Questions for Class 12 Chemistry (15)
• MCQ Questions for Class 12 Chemistry (8)
• NCERT Solutions for Class 12 Chemistry (16)
• Revision Notes for Class 12 Chemistry (7)
• Assertion Reason Questions for Class 12 Economics (9)
• Case Study Questions for Class 12 Economics (9)
• MCQ Questions for Class 12 Economics (1)
• MCQ Questions for Class 12 English (2)
• Assertion Reason Questions for Class 12 Entrepreneurship (7)
• Case Study Questions for Class 12 Entrepreneurship (7)
• Case Study Questions for Class 12 Geography (18)
• Assertion Reason Questions for Class 12 History (8)
• Case Study Questions for Class 12 History (13)
• Assertion Reason Questions for Class 12 Informatics Practices (13)
• Case Study Questions for Class 12 Informatics Practices (11)
• MCQ Questions for Class 12 Informatics Practices (5)
• Assertion and Reason Questions for Class 12 Maths (14)
• Case Study Questions for Class 12 Maths (13)
• Maths Formulas for Class 12 (5)
• MCQ Questions for Class 12 Maths (14)
• Problems Based on Class 12 Maths (1)
• RD Sharma Solutions for Class 12 Maths (1)
• Assertion Reason Questions for Class 12 Physical Education (11)
• Case Study Questions for Class 12 Physical Education (11)
• MCQ Questions for Class 12 Physical Education (10)
• Assertion Reason Questions for Class 12 Physics (16)
• Case Study Based Questions for Class 12 Physics (14)
• Class 12 Physics Conceptual Questions (16)
• Class 12 Physics Discussion Questions (1)
• Class 12 Physics Latest Updates (2)
• Derivations for Class 12 Physics (8)
• Extra Questions for Class 12 Physics (4)
• Important Questions for Class 12 Physics (8)
• MCQ Questions for Class 12 Physics (14)
• NCERT Solutions for Class 12 Physics (18)
• Numerical Problems Based on Class 12 Physics (16)
• Physics Class 12 Viva Questions (1)
• Revision Notes for Class 12 Physics (7)
• Assertion Reason Questions for Class 12 Political Science (16)
• Case Study Questions for Class 12 Political Science (16)
• Notes for Class 12 Political Science (1)
• Assertion Reason Questions for Class 6 Maths (13)
• Case Study Questions for Class 6 Maths (13)
• Extra Questions for Class 6 Maths (1)
• Worksheets for Class 6 Maths (1)
• Assertion Reason Questions for Class 6 Science (16)
• Case Study Questions for Class 6 Science (16)
• Extra Questions for Class 6 Science (1)
• MCQ Questions for Class 6 Science (9)
• Assertion Reason Questions for Class 6 Social Science (1)
• Case Study Questions for Class 6 Social Science (26)
• NCERT Exemplar for Class 7 Maths (13)
• NCERT Exemplar for Class 7 Science (19)
• NCERT Exemplar Solutions for Class 7 Maths (12)
• NCERT Exemplar Solutions for Class 7 Science (18)
• NCERT Notes for Class 7 Science (18)
• Assertion Reason Questions for Class 7 Maths (14)
• Case Study Questions for Class 7 Maths (14)
• Extra Questions for Class 7 Maths (5)
• Assertion Reason Questions for Class 7 Science (18)
• Case Study Questions for Class 7 Science (17)
• Extra Questions for Class 7 Science (19)
• Assertion Reason Questions for Class 7 Social Science (1)
• Case Study Questions for Class 7 Social Science (30)
• Assertion Reason Questions for Class 8 Maths (7)
• Case Study Questions for Class 8 Maths (17)
• Extra Questions for Class 8 Maths (1)
• MCQ Questions for Class 8 Maths (6)
• Assertion Reason Questions for Class 8 Science (16)
• Case Study Questions for Class 8 Science (11)
• Extra Questions for Class 8 Science (2)
• MCQ Questions for Class 8 Science (4)
• Numerical Problems for Class 8 Science (1)
• Revision Notes for Class 8 Science (11)
• Assertion Reason Questions for Class 8 Social Science (27)
• Case Study Questions for Class 8 Social Science (23)
• CBSE Class 9 English Beehive Notes and Summary (2)
• Assertion Reason Questions for Class 9 Maths (14)
• Case Study Questions for Class 9 Maths (14)
• MCQ Questions for Class 9 Maths (11)
• NCERT Notes for Class 9 Maths (6)
• NCERT Solutions for Class 9 Maths (12)
• Revision Notes for Class 9 Maths (3)
• Study Notes for Class 9 Maths (10)
• Assertion Reason Questions for Class 9 Science (16)
• Case Study Questions for Class 9 Science (15)
• Evergreen Science Book Solutions for Class 9 (15)
• Extra Questions for Class 9 Science (22)
• MCQ Questions for Class 9 Science (11)
• NCERT Solutions for Class 9 Science (15)
• Revision Notes for Class 9 Science (1)
• Study Notes for Class 9 Science (15)
• Topic wise MCQ Questions for Class 9 Science (2)
• Topicwise Questions and Answers for Class 9 Science (15)
• Assertion Reason Questions for Class 9 Social Science (15)
• Case Study Questions for Class 9 Social Science (19)
• CHEMISTRY (8)
• Chemistry Articles (2)
• Daily Practice Problems (DPP) (3)
• Books for CBSE Class 9 (1)
• Books for ICSE Class 10 (3)
• Editable Study Materials (8)
• Exam Special for CBSE Class 10 (3)
• H. C. Verma (Concepts of Physics) (13)
• Study Materials for ICSE Class 10 Biology (14)
• Extra Questions for ICSE Class 10 Chemistry (1)
• Study Materials for ICSE Class 10 Chemistry (5)
• Study Materials for ICSE Class 10 Maths (16)
• Important Questions for ICSE Class 10 Physics (13)
• MCQ Questions for ICSE Class 10 Physics (4)
• Study Materials for ICSE Class 10 Physics (8)
• Study Materials for ICSE Class 9 Maths (7)
• Study Materials for ICSE Class 9 Physics (10)
• Topicwise Problems for IIT Foundation Mathematics (4)
• Challenging Physics Problems for JEE Advanced (2)
• Topicwise Problems for JEE Physics (1)
• DPP for JEE Main (1)
• Integer Type Questions for JEE Main (1)
• Integer Type Questions for JEE Chemistry (6)
• Chapterwise Questions for JEE Main Physics (1)
• Integer Type Questions for JEE Main Physics (8)
• Physics Revision Notes for JEE Main (4)
• JEE Mock Test Physics (1)
• JEE Study Material (1)
• JEE/NEET Physics (6)
• CBSE Syllabus (1)
• Maths Articles (2)
• NCERT Books for Class 12 Physics (1)
• NEET Chemistry (13)
• Important Questions for NEET Physics (17)
• Topicwise DPP for NEET Physics (5)
• Topicwise MCQs for NEET Physics (32)
• NTSE MAT Questions (1)
• Physics (1)
• Alternating Current (1)
• Electrostatics (6)
• Fluid Mechanics (2)
• PowerPoint Presentations (13)
• Previous Years Question Paper (3)
• Products for CBSE Class 10 (15)
• Products for CBSE Class 11 (10)
• Products for CBSE Class 12 (6)
• Products for CBSE Class 6 (2)
• Products for CBSE Class 7 (5)
• Products for CBSE Class 8 (1)
• Products for CBSE Class 9 (3)
• Products for Commerce (3)
• Products for Foundation Courses (2)
• Products for JEE Main & Advanced (10)
• Products for NEET (6)
• Products for ICSE Class 6 (1)
• Electrostatic Potential and Capacitance (1)
• Topic Wise Study Notes (Physics) (2)
• Topicwise MCQs for Physics (2)
• Uncategorized (138)

Test series for students preparing for Engineering & Medical Entrance Exams are available. We also provide test series for School Level Exams. Tests for students studying in CBSE, ICSE or any state board are available here. Just click on the link and start test.

Download CBSE Books

Exam Special Series:

• Sample Question Paper for CBSE Class 10 Science (for 2024)
• Sample Question Paper for CBSE Class 10 Maths (for 2024)
• CBSE Most Repeated Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Important Numericals Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
• Sample Question Papers for CBSE Class 12 Physics (for 2024)
• Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
• Sample Question Papers for CBSE Class 12 Maths (for 2024)
• Sample Question Papers for CBSE Class 12 Biology (for 2024)
• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• 10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

Leave a Reply Cancel reply

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

• Andhra Pradesh
• Chhattisgarh
• West Bengal
• Madhya Pradesh
• Maharashtra
• Jammu & Kashmir
• NCERT Books 2022-23
• NCERT Solutions
• NCERT Notes
• NCERT Exemplar Books
• NCERT Exemplar Solution
• States UT Book
• School Kits & Lab Manual
• NCERT Books 2021-22
• NCERT Books 2020-21
• NCERT Book 2019-2020
• NCERT Book 2015-2016
• RD Sharma Solution
• TS Grewal Solution
• TR Jain Solution
• Selina Solution
• Frank Solution
• Lakhmir Singh and Manjit Kaur Solution
• I.E.Irodov solutions
• ICSE - Goyal Brothers Park
• ICSE - Dorothy M. Noronhe
• Micheal Vaz Solution
• S.S. Krotov Solution
• Evergreen Science
• KC Sinha Solution
• ICSE - ISC Jayanti Sengupta, Oxford
• ICSE Focus on History
• ICSE GeoGraphy Voyage
• ICSE Hindi Solution
• ICSE Treasure Trove Solution
• Thomas & Finney Solution
• SL Loney Solution
• SB Mathur Solution
• P Bahadur Solution
• Narendra Awasthi Solution
• MS Chauhan Solution
• LA Sena Solution
• Integral Calculus Amit Agarwal Solution
• IA Maron Solution
• Hall & Knight Solution
• Errorless Solution
• Pradeep's KL Gogia Solution
• OP Tandon Solutions
• Sample Papers
• Previous Year Question Paper
• Important Question
• Value Based Questions
• CBSE Syllabus
• CBSE MCQs PDF
• Assertion & Reason
• New Revision Notes
• Revision Notes
• Question Bank
• Marks Wise Question
• Toppers Answer Sheets
• Exam Paper Aalysis
• Concept Map
• CBSE Text Book
• Additional Practice Questions
• Vocational Book
• CBSE - Concept
• KVS NCERT CBSE Worksheets
• Formula Class Wise
• Formula Chapter Wise
• Toppers Notes
• Most Repeated Question
• Diagram Based Question
• Study Planner
• JEE Previous Year Paper
• JEE Mock Test
• JEE Crash Course
• JEE Sample Papers
• Important Info
• SRM-JEEE Previous Year Paper
• SRM-JEEE Mock Test
• VITEEE Previous Year Paper
• VITEEE Mock Test
• BITSAT Previous Year Paper
• BITSAT Mock Test
• Manipal Previous Year Paper
• Manipal Engineering Mock Test
• AP EAMCET Previous Year Paper
• AP EAMCET Mock Test
• COMEDK Previous Year Paper
• COMEDK Mock Test
• GUJCET Previous Year Paper
• GUJCET Mock Test
• KCET Previous Year Paper
• KCET Mock Test
• KEAM Previous Year Paper
• KEAM Mock Test
• MHT CET Previous Year Paper
• MHT CET Mock Test
• TS EAMCET Previous Year Paper
• TS EAMCET Mock Test
• WBJEE Previous Year Paper
• WBJEE Mock Test
• AMU Previous Year Paper
• AMU Mock Test
• CUSAT Previous Year Paper
• CUSAT Mock Test
• AEEE Previous Year Paper
• AEEE Mock Test
• UPSEE Previous Year Paper
• UPSEE Mock Test
• CGPET Previous Year Paper
• BCECE Previous Year Paper
• JCECE Previous Year Paper
• Crash Course
• Previous Year Paper
• NCERT Based Short Notes
• NCERT Based Tests
• NEET Sample Paper
• Previous Year Papers
• Quantitative Aptitude
• Numerical Aptitude Data Interpretation
• General Knowledge
• Mathematics
• Agriculture
• Accountancy
• Business Studies
• Political science
• Enviromental Studies
• Mass Media Communication
• Teaching Aptitude
• Verbal Ability & Reading Comprehension
• Logical Reasoning & Data Interpretation
• CAT Mock Test
• CAT Important Question
• CAT Vocabulary
• CAT English Grammar
• MBA General Knowledge
• CAT Mind Map
• CAT Study Planner
• CMAT Mock Test
• SRCC GBO Mock Test
• SRCC GBO PYQs
• XAT Mock Test
• SNAP Mock Test
• IIFT Mock Test
• MAT Mock Test
• CUET PG Mock Test
• CUET PG PYQs
• MAH CET Mock Test
• MAH CET PYQs
• NAVODAYA VIDYALAYA
• SAINIK SCHOOL (AISSEE)
• Mechanical Engineering
• Electrical Engineering
• Electronics & Communication Engineering
• Civil Engineering
• Computer Science Engineering
• CBSE Board News
• Scholarship Olympiad
• School Admissions
• Entrance Exams
• All Board Updates
• Miscellaneous
• State Wise Books
• Engineering Exam

SHARING IS CARING If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
• CBSE Syllabus 2023-24
• Social Media Channels
• Login Customize Your Notification Preferences
• Integrals 3 September, 2024, 11:00 am
• Inverse Trigonometric Functions 3 September, 2024, 11:00 am
• Relations And Functions 14 April, 2021, 6:25 pm
• Matrices 14 April, 2021, 6:25 pm
• Determinants 14 April, 2021, 6:25 pm
• Continuity And Differentiability 14 April, 2021, 6:25 pm
• Application Of Derivatives 14 April, 2021, 6:25 pm
• Application Of Integrals 14 April, 2021, 6:25 pm
• Differential Equations 14 April, 2021, 6:25 pm

• Second click on the toggle icon

Provide prime members with unlimited access to all study materials in PDF format.

Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding.

Provide prime users with access to exclusive PDF study materials that are not available to regular users.

• Class 6 Maths
• Class 6 Science
• Class 6 Social Science
• Class 6 English
• Class 7 Maths
• Class 7 Science
• Class 7 Social Science
• Class 7 English
• Class 8 Maths
• Class 8 Science
• Class 8 Social Science
• Class 8 English
• Class 9 Maths
• Class 9 Science
• Class 9 Social Science
• Class 9 English
• Class 10 Maths
• Class 10 Science
• Class 10 Social Science
• Class 10 English
• Class 11 Maths
• Class 11 Computer Science (Python)
• Class 11 English
• Class 12 Maths
• Class 12 English
• Class 12 Economics
• Class 12 Accountancy
• Class 12 Physics
• Class 12 Chemistry
• Class 12 Biology
• Class 12 Computer Science (Python)
• Class 12 Physical Education
• GST and Accounting Course
• Excel Course
• Tally Course
• Finance and CMA Data Course
• Payroll Course

Interesting

• Learn English
• Learn Excel
• Learn Tally
• Learn GST (Goods and Services Tax)
• Learn Accounting and Finance
• GST Tax Invoice Format
• Accounts Tax Practical
• Tally Ledger List
• GSTR 2A - JSON to Excel

Are you in school ? Do you love Teachoo?

We would love to talk to you! Please fill this form so that we can contact you

Case Based Questions (MCQ)

• NCERT Exemplar - MCQs
• Forming equations and solving
• Questions - Forming equations and solving

Question 1 - Case Based Questions (MCQ) - Chapter 12 Class 12 Linear Programming

Last updated at April 16, 2024 by Teachoo

An aeroplane can carry a maximum of 200 passengers.  A profit of Rs. 1000 is made on each executive class  ticket and a profit of Rs. 600 is made on each economy  class ticket. The airline reserves at least 20 seats for  the executive class. However, at least 4 times as many  passengers prefer to travel by economy class, than by  executive class. It is given that the number of executive class tickets is x and that of economy class tickets is y.

This question is inspired from Misc 5 - Chapter 12 Class 12 (Linear Programming)

The maximum value of x + y is_______.

(a) 100  , (c) 20  .

The relation between x and y is_______.

(a) x < y , (b) y > 80, (c) x ≥ 4y      , (d) y ≥ 4x.

Which among these is not a constraint for this LPP?

(a) x ≥ 0   , (b) x + y ≤ 200, (c) x ≥ 80   , (d) 4x − y ≤ 0.

The profit when x = 20 and y = 80 is  _______.

(a) rs. 60000  , (b) rs. 68000, (c) rs. 64000  , (d) rs. 136000.

The maximum profit is Rs.  _______.

(a) 136000 , (c) 68000 , (d) 120000 .

Question An aeroplane can carry a maximum of 200 passengers. A profit of Rs. 1000 is made on each executive class ticket and a profit of Rs. 600 is made on each economy class ticket. The airline reserves at least 20 seats for the executive class. However, at least 4 times as many passengers prefer to travel by economy class, than by executive class. It is given that the number of executive class tickets is x and that of economy class tickets is y. Let the Number of Executive Class tickets = x Number of Economy Class tickets = y Given that Aeroplane can carry a maximum of 200 passengers. ∴ x + y ≤ 200 Also, The airline reserves at least 20 seats for the executive class. ∴ x ≥ 20 And, at least 4 times as many passengers prefer to travel by economy class, than by executive class y ≥ 4x And, y ≥ 0 We would need to Maximize Profit Given that A profit of Rs. 1000 is made on each executive class ticket and a profit of Rs. 600 is made on each economy class ticket Z = 1000x + 600y Combining all constraints : Maximize Z = 1000x + 600y Subject to Constraints : x + y ≤ 200 y ≥ 4x x ≥ 20 y ≥ 0 Question 1 The maximum value of x + y is_______. (a) 100 (b) 200 (c) 20 (d) 80 Since x + y ≤ 200 So, the correct answer is (B) Question 2 The relation between x and y is_______. (a) x < y (b) y > 80 (c) x ≥ 4y (d) y ≥ 4x From our constraints y ≥ 4x So, the correct answer is (D) Question 3 Which among these is not a constraint for this LPP? (a) x ≥ 0 (b) x + y ≤ 200 (c) x ≥ 80 (d) 4x − y ≤ 0 Constraints for this Linear Programming Problem (LPP) are x + y ≤ 200 y ≥ 4x x ≥ 20 y ≥ 0 Since x ≥ 80 is not in the constraints So, the correct answer is (C) Question 4 The profit when x = 20 and y = 80 is _______. (a) Rs. 60000 (b) Rs. 68000 (c) Rs. 64000 (d) Rs. 136000 Now, Z = 1000x + 600y Putting x = 20, y = 80 Z = 1000(20) + 600(80) Z = 20,000 + 48,000 Z = 68,000 ∴ Profit = Rs 68,000 So, the correct answer is (B) Question 5 The maximum profit is Rs. _______. (a) 136000 (b) 128000 (c) 68000 (d) 120000 Our LPP is Maximize Z = 1000x + 600y Subject to Constraints : x + y ≤ 200 y ≥ 4x x ≥ 20, y ≥ 0 Hence, Profit will be maximum, if Tickets of executive class = 40 Tickets of economy class = 160 Maximum Profit = Rs 1,36,000 So, the correct answer is (a)

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Please login to view more pages. it's free :), solve all your doubts with teachoo black.

• Neet Online Test Pack

12th Standard stateboard question papers & Study material

தமிழ் subjects.

கணினி பயன்பாடுகள்

கணினி அறிவியல்

வணிகக் கணிதம் மற்றும் புள்ளியியல்.

கணினி தொழில்நுட்பம்

கணக்குப்பதிவியல்

English Subjects

Computer Science

Business Maths and Statistics

Accountancy

Computer Applications

Computer Technology

11th Standard stateboard question papers & Study material

9th Standard stateboard question papers & Study material

Social Science

சமூக அறிவியல்

6th standard stateboard question papers & study material.

10th Standard stateboard question papers & Study material

7th Standard stateboard question papers & Study material

8th Standard stateboard question papers & Study material

கணிதம் - old

12th Standard CBSE Subject Question Paper & Study Material

Introductory Micro and Macroeconomics

Business Studies

Indian Society

Physical Education

Bio Technology

Engineering Graphics

Entrepreneurship

Hindi Elective

Home Science

Legal Studies

Political Science

11th Standard CBSE Subject Question Paper & Study Material

Mathematics

Enterprenership

Applied Mathematics

10th standard cbse subject question paper & study material.

9th Standard CBSE Subject Question Paper & Study Material

8th Standard CBSE Subject Question Paper & Study Material

7th Standard CBSE Subject Question Paper & Study Material

6th Standard CBSE Subject Question Paper & Study Material

School Exams

Tamil Nadu State Board Exams

Scholarship Exams

Study Materials , News and Scholarships

Stateboard Tamil Nadu

Free Online Tests

Educational News

Scholarships

Entrance Exams India

Video Materials

12th Standard CBSE

CBSE 12th Standard Maths Subject Linear Programming Case Study Questions With Solution 2021

CBSE 12th Standard Maths Subject Linear Programming Case Study Questions With Solution 2021 Shalini Sharma - Udaipur May-21 , 2021

QB365 Provides the updated CASE Study Questions for Class 12 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

A PHP Error was encountered

Severity: Warning

Message: in_array() expects parameter 2 to be array, null given

Filename: material/details.php

Line Number: 1436

Message: Use of undefined constant EXAM - assumed 'EXAM' (this will throw an Error in a future version of PHP)

Line Number: 1438

QB365 - Question Bank Software

Cbse 12th standard maths linear programming case study questions with solution 2021.

Final Semester - June 2015

Case Study Questions

Linear programming is a method for finding the optimal values (maximum or minimum) of quantities subject to the constraints when relationship is expressed as linear equations or inequations. Based on the above information, answer the following questions. (i) The optimal value of the objective function is attained at the points

(ii) The graph of the inequality 3x + 4y < 12 is

(iv) The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is

(v)  The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20,40), (60,20), (60, 0). The objective function is Z = 4x + 3y. Compare the quantity in Column A and Column B

(ii) The corner points of the feasible region shown in above graph are

(iii) If Z = x + y be the objective function and max Z = 30. The maximum value occurs at point

(iv) If Z = 6x - 9y be the objective function, then maximum value of Z is

(v) If Z = 6x + 3y be the objective function, then what is the minimum value of Z?

Corner points of the feasible region for an LPP are (0, 3), (5, 0), (6, 8), (0, 8). Let Z = 4x - 6y be the objective function. Based on the above information, answer the following questions. (i) The minimum value of Z occurs at

(ii) Maximum value of Z occurs at

(iii) Maximum of Z - Minimumof Z =

(v) The feasible solution of LPP belongs to

(ii) Let the constraints in the given problem is represented by the following inequalities x + y ≤ 20 360x + 240y ≤ 5760 x, y ≥ 0 Then which of the following point lie in its feasible region.

(iii) If the objective function of the given problem is maximise z = 22x + 18y, then its optimal value occur at

(iv) Suppose the following shaded region APDO, represent the feasible region corresponding to mathematical formulation. of given problem. Then which of the following represent the coordinates of one of its corner points.

Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to, constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region. Based on the above information, answer the following questions. (i) Objective function of a L.P.P. is

(ii) Which of the following statement is correct?

(iii) In solving the LPP : "minimize f = 6x + 10y subject to constraints x ≥  6, Y ≥ 2, 2x + y  ≥ 10,x  ≥ 0,y ≥ 0" redundant constraints are

*****************************************

• Previous 12th Maths Vector Algebra Chapter Case Study Question with Answers CBSE
• Next 12th Maths Three Dimensional Geometry Chapter Case Study Question with Answers C...

Reviews & Comments about CBSE 12th Standard Maths Subject Linear Programming Case Study Questions With Solution 2021

Write your Comment

12th Standard CBSE Maths Videos

CBSE 12th Maths Sample Model Question Paper with Answer Key 2023

12th Standard CBSE Maths Usefull Links

• 10th Standard

Other 12th Standard CBSE Subjects

Other 12th Standard CBSE Maths Study material

12th maths vector algebra chapter case study question with answers cbse click to view, 12th maths three dimensional geometry chapter case ... click to view, 12th maths probability chapter case study question with answers cbse click to view, 12th maths linear programming chapter case study question with answers cbse click to view, 12th maths differential equations chapter case study question with answers cbse click to view, 12th maths continuity and differentiability chapter case ... click to view, 12th maths application of integrals chapter case ... click to view, cbse 12th standard maths subject differential equations ... click to view, cbse 12th standard maths subject determinants value ... click to view, cbse 12th standard maths subject value based ... click to view, register & get the solution for cbse 12th standard maths subject linear programming case study questions with solution 2021.

Class 12 Maths: Case Study Based Questions PDF Download

• Post author: studyrate
• Post published:
• Post category: 12 board / Class 12
• Post comments: 0 Comments

You must practice some good Case Study questions of Class 12 Maths to boost your preparation to score 95+% on Boards. In this post, you will get Case Study Questions of All Chapters which will come in CBSE Class 12 Maths Board Exams.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

We have provided here Case Study questions for the Class 12 Maths exams. You can read these chapter-wise Case Study questions. Prepared by subject experts and experienced teachers. The answer key is also provided so that you can check the correct answer for each question. Practice these questions to score well in your Board Final exams.

We are providing Case Study questions for class 12 Biology based on the latest syllabi. There is a total of 13 chapters included in CBSE class 12 Maths exams. Students can practice these questions for concept clarity and score better marks in their exams.

Table of Contents

CBSE Class 12th – MATHS : Chapterwise Case Study Question & Solution

CBSE will ask two Case Study Questions in the CBSE class 12 maths questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them.

Case Study Based Questions for Class 12 Maths

Class 12 Physics Case Study Questions Class 12 Chemistry Case Study Questions Class 12 Biology Case Study Questions Class 12 Maths Case Study Questions

Books for Class 12 Maths

Strictly as per the new term-wise syllabus for Board Examinations to be held in the academic session 2022-23 for class 12 Multiple Choice Questions based on new typologies introduced by the board- Stand-Alone MCQs, MCQs based on Assertion-Reason Case-based MCQs. Include Questions from CBSE official Question Bank released in April 2022 Answer key with Explanations What are the updates in the book: Strictly as per the Term wise syllabus for Board Examinations to be held in the academic session 2022-23. Chapter-wise -Topic-wise Multiple choice questions based on the special scheme of assessment for Board Examination for Class 12th.

Class 12 Maths Syllabus 2022-23

Unit-I: Relations and Functions

1. Relations and Functions (15 Periods)

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions.

2. Inverse Trigonometric Functions (15 Periods)

Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit-II: Algebra

1. Matrices (25 Periods)

Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operation on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Oncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants 25 Periods

Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit-III: Calculus

1. Continuity and Differentiability (20 Periods)

Continuity and differentiability, chain rule, derivative of inverse trigonometric functions, 𝑙𝑖𝑘𝑒 sin −1  𝑥 , cos −1  𝑥 and tan −1  𝑥, derivative of implicit functions. Concept of exponential and logarithmic functions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.

2. Applications of Derivatives (10 Periods)

Applications of derivatives: rate of change of bodies, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as reallife situations).

3. Integrals (20 Periods)

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals (15 Periods)

Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only)

5. Differential Equations (15 Periods)

Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

Unit-IV: Vectors and Three-Dimensional Geometry

1. Vectors (15 Periods)

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors.

2. Three – dimensional Geometry (15 Periods)

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines.

Unit-V: Linear Programming

1. Linear Programming (20 Periods)

Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Unit-VI: Probability

1. Probability 30 (Periods)

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

You Might Also Like

Lost spring summary class 12 english pdf, cbse class 12 physics term 1 mcq questions with explanations pdf download, class 12th english chapters summaries flamingo and vistas, leave a reply cancel reply.

Save my name, email, and website in this browser for the next time I comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed .

The Topper Combo Flashcards

• Contains the Latest NCERT in just 350 flashcards.
• Colourful and Interactive
• Summarised Important reactions according to the latest PYQs of NEET(UG) and JEE

No thanks, I’m not interested!

CBSE Class 12 Maths –Chapter 12 Linear Programming- Study Materials

NCERT Solutions Class 12 All Subjects Sample Papers Past Years Papers

Notes and Study Materials

• Concepts of Linear Programming
• Master File for Linear Programming
• Linear Programming Note
• NCERT Solutions for – Linear Programming
• NCERT Exemplar Solutions for – Linear Programming
• R D Sharma Solution of Linear Programming
• Past Many Years CBSE Questions and Answer Of Relation and Function
• Linear Programming Mind Map

Examples and Exercise

• Linear Programming : Practice Paper 1
• Linear Programming : Practice Paper 2
• Linear Programming : Practice Paper 3
• Linear Programming : Practice Paper 4

CBSE Revision Notes for CBSE Class 12 Mathematics Linear Programming Introduction, related terminology such as constraints, objective function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions(bounded and unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Linear Programming & Its Applications 

• It is an important optimization (maximization or minimization) technique used in decision making is business and everyday life for obtaining the maximum or minimum values as required of a linear expression to satisfying certain number of given linear restrictions.

• Involve finding maximum profit, minimum cost, or minimum use of resources etc.

• Used in industry, commerce, management science etc.

Linear Programming Problem and its Mathematical Formulation:  Definitions 

Optimal value • Maximum or Minimum value of a linear function

Objective Function • The function which is to be optimized (maximized/minimized) • Linear function Z = ax + by, where a, b are constants, which has to be maximised or minimized is called a linear objective function. • Eg:Z = 250x+ 75y where variables x and y are called decision variables

Linear Constraints   • System of linear inequations/equations under which objective function is to be optimized • Linear inequalities/equations or restrictions on the variables of a linear programming problem • Also called as Overriding Conditions or Constraints • The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.

Non-negative Restrictions • All the variables considered for making decisions assume non-negative values.

Optimisation problem •  A problem which seeks to maximise or minimise a linear function (say of two variables x and y) subject to certain constraints as determined by a set of linear inequalities • Linear programming problems (LPP) are special type of optimisation problems.

Note:  • The term linear implies that all the mathematical relations used in the problem are linear relations . • The term programming refers to the method of determining a particular programme or plan of action.

Mathematical Formulation of the Problem 

A general LPP can be stated as (Max/Min) Z- c₁x₁ + c₂x₂ + CₙXₙ (Objective function) subject to constraints and the non-negative restriction

• x₁, X₂,….., Xₙ ≥ 0 where all a₁₁, a₁₂, ……, amn; • b₁, b₂…….bm; C₁, C₂ Cₙ are constants and • X₁, X₂. ……. Xₙ are variables

Graphical method of solving linear programming problems  Terminologies 

Solution of a LPP • A set of values of the variables x₁, X₂,….., Xₙ satisfying the constraints of a LPP

Feasible Solution of a LPP • A set of values of the variables x₁, X₂,….., Xₙ satisfying the constraints and non-negative restrictions of a LPP • Points within and on the boundary of the feasible region represent feasible solutions of the constraints

Feasible Region  • The common region determined by all the constraints including non-negative constraints x, y ≥ 0 of a linear programming problem is called the feasible region (or solution region) for the problem

Feasible Choice  • Each point in the Feasible region

Infeasible Region  • Region outside the feasible region

Infeasible Solution  • Any point outside the feasible region Optimal Solution of a LPP • A feasible solution of a LPP is said to, be optimal (or optimum), if it also optimizes the objective f unction of the problem. Graphical Solution of a LPP  • The solution of a LPP obtained by graphical method ie.., by drawing the graphs corresponding to the constraints and the non-negative restrictions

Unbounded Solution • If the value of the objective function can be increased or decreased indefinitely, such solutions

Graph the constraints stated as linear inequalities:

5x + y ≥ 100…………….. Equation (1) x+y ≥ 60…………………… Equation (2) x ≥ 0…………………………..Equation (3) y ≥ 0…………………………..Equation (4)

For Plotting the Equation (1),

• Let x = 0, Hence we get the point y=100 • Let y = 0, Hence we get the point x-100/5 – 20 • Equation (1) is obtained by joining the points (20,100)

For Plotting the Equation (2),

• Let x = 0, Hence we get the point y=60 • Let y = 0, Hence we get the point x=60 • Equation (2) is obtained by joining the points (60,60)

From Equation (3) and Equation (4) we know both x and y are greater than 0

From the graph above,

• Points within and on the boundary of the feasible region represent feasible solutions of the constraints.

• In Fig 12.1, every point within and on the boundary of the feasible region OABC represents feasible solution to the problem.

• For example, the point (10, 50) is a feasible solution of the problem and so are the points (0, 60), (20, 0) etc.

• Any point outside the feasible region is called an infeasible solution. For example, the point (25, 40) is an infeasible solution of the problem.

• Now, we see that every point in the feasible region OABC satisfies all the constraints as given in (1) to (4), and since there are infinitely many points, it is not evident how we should go about finding a point that gives a maximum value of the objective function Z= 250x + 75y

• Let R be the feasible region (convex polygon) for a linear programming problem • Let Z= ax + by be the objective function. • When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region.

• Let R be the feasible region for a linear programming problem • Let Z = ax + by be the objective function. • If R is bounded**, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R. •Remark • If R is unbounded, then a maximum or a minimum value of the objective function may not exist. However, if it exists, it must occur at a corner point of R .

Graphical Method to Solve a Linear Programming Problem

• There are two techniques of solving a LPP by graphical method

1. Corner point method 2. Iso-profit or Iso-cost method

Corner Point Method 

• Based on the principle of extreme point theorem. • Procedure to Solve a LPP Graphically by Corner Point Method • Consider each constraint as an equation. • Plot each equation on graph, as each one will geometrically represent a straight line. • The common region, thus obtained satisfying all the constraints and the non-negative restrictions is called the feasible region. It is a convex polygon . • Determine the vertices (corner points) of the convex polygon. These vertices are known as the extreme points of corners of the feasible region. • Find the values of the objective function at each of the extreme points. • The point at which the value of the objective function is optimum (maximum or minimum) is the optimal solution of the given LPP.

Isom-profit or Iso-cost Method 

Procedure to Solve a LPP Graphically by Iso-profit or Iso-cost Method

• Consider each constraint as an equation. • Plot each equation on graph as each one will geometrically represent a straight line. • The polygonal region so obtained, satisfying all the constraints and the non-negative restrictions is the convex set of all feasible solutions of the given LPP, which is also known as feasible region. • Determine the extreme points of the feasible region. • Give some convenient value k to the objective function Z and draw the corresponding straight line in the xy-plane. • If the problem is of maximization, then draw lines parallel to the line Z = k and obtain a line which is farthest from the origin and has atleast one point common to the feasible region. • If the problem is of minimization, then draw lines parallel to the line Z- k and obtain a line, which is nearest to the origin and has atleast one point common to the feasible region. • The common point so obtained is the optimal solution of the given LPP. Working Rule for Marking Feasible Region 

Consider the constraint ax + by ≥ C, where c > 0.

• First draw the straight line ax + by = c by joining any two points on it. • For this find two convenient points satisfying this equation. • This straight line divides the xy-plane in two parts. • The inequation ax + by c will represent that part of the xy-plane which lies to that side of the line ax + by = c in which the origin lies.

Again, consider the constraint ax + by ≥ c, where c > 0.

• Draw the straight line ax + by = c by joining any two points on it. • This straight line divides the xy-plane in two parts. • The inequation ax + by ≥ c will represent that part of the xy-plane, which lies to that side of the line ax + by = c in which the origin does not lie.

Important Points to be remembered 

1. Basic Feasible Solution • A BFS is a basic solution which also satisfies the non-negativity restrictions.

2. Optimum Basic Feasible Solution • A BFS is said to be optimum, if it also optimizes (Max or min) the objective function.

Solve the following linear programming problem graphically: Maximise Z- 4x + y…………………………… Equation (1)

Subject to the constraints:

• x + y ≤ 50……………………Equation (2) • 3x + y ≤ 90………………….Equation (3) • x ≥ 0 , y ≥ 0…………………Equation (4)

• Let x = 0, Hence we get the point y=50 • Let y = 0, Hence we get the point x= 50 • Equation (2) is obtained by joining the points (50,50)

For Plotting the Equation (3),

• Let x = 0, Hence we get the point y=90 • Let y = 0, Hence we get the point x=30 • Equation (3) is obtained by joining the points (30,90)

From Equation (4) we know both x and y are greater than 0

Hence the points are (0, 0), (50, 50), (0, 50), (30, 0), (30, 90). The shaded region in graph is the feasible region determined by the system of constraints (2) to (4). We observe that the feasible region OABC is bounded.

Using the Corner Point Method to determine the maximum value of Z by substituting the vertices of the bounded region. Hence, maximum value of Z is 120 at the point (30, 0).

Determine graphically the minimum value of the objective function

Z = – 50x + 20y …..(1)

• 2x -y ≥ -5…… (2) • 3x + y ≥ 3….. (3) • 2x – 3y ≤ 12.. (4) • x ≥ 0, y ≥ 0…. (5)

Graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown.

By Observation that the feasible region is unbounded.

We now evaluate Z at the corner points.

From this table, we find that – 300 is the smallest value of Z at the corner point (6, 0).

Since the region would have been bounded, this smallest value of Z is the minimum value of Z

(Theorem 2). 

But here we see that the feasible region is unbounded. Therefore, – 300 may or may not be the minimum value of Z.

To decide this issue, we graph the inequality 50x + 20y < 300

i.e., 5x + 2y < 30 (By dividing the Equation above by 10)

And check whether the resulting open half plane has points in common with feasible region or not.

If it has common points, then -300 will not be the minimum value of Z. Otherwise, -300 will be the minimum value of Z.

As shown in the graph above, it has common points. Therefore, Z= -50 x + 20 y has no minimum value subject to the given constraints.

General features of linear programming problems 

1. The feasible region is always a convex region. 2. The maximum (or minimum) solution of the objective function occurs at the vertex (corner) of the feasible region. 3. If two corner points produce the same maximum (or minimum) value of the objective function, then every point on the line segment joining these points will also give the same maximum (or minimum) value.

Different Types of Linear Programming Problems 

A few important linear programming problems are listed below:

1. Manufacturing problems

• Determine the number of units of different products which should be produced and sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output etc., in order to make maximum profit.

2. Diet problems

• Determine the amount of different kinds of constituents/nutrients which should be included in a diet so as to minimise the cost of the desired diet such that it contains a certain minimum amount of each constituent/nutrients.

3. Transportation problems

• Determine a transportation schedule in order to find the cheapest way of transporting a product from plants/factories situated at different locations to different markets.

Class 12 CBSE Applied Maths ML AGGARWAL Linear Programming Case Study

Please select.

View Solution  

The Site is down as we are performing important server maintenance, during which time the server will be unavailable for approximately 24 hours. Please hold off on any critical actions until we are finished. As always your feedback is appreciated.

• Study Packages
• NCERT Solutions
• Sample Papers
• Online Test

• Questions Bank
• Mathematics
• Linear Programming
• Test Series
• Ncert Solutions
• Solved Papers
• Current Affairs
• JEE Main & Advanced
• Pre-Primary
• MP State Exams
• UP State Exams
• Rajasthan State Exams
• Jharkhand State Exams
• Chhattisgarh State Exams
• Bihar State Exams
• Haryana State Exams
• Gujarat State Exams
• MH State Exams
• Himachal State Exams
• Delhi State Exams
• Uttarakhand State Exams
• Punjab State Exams
• J&K State Exams

12th Class Mathematics Linear Programming Question Bank

Done case based (mcqs) - linear programming total questions - 25.

A) on X-axis done clear

B) on Y-axis done clear

C) which are corner points of the feasible region done clear

D) None of these done clear

question_answer 2) The graph of the inequality \[3x+\text{4}y<12\] is

A) half plane that contains the origin done clear

B) half plane that neither contains the origin nor the points of the line \[3x+4y=12\]. done clear

C) Whole XOY-plane excluding the points on line \[3x+4y=12\] done clear

D) None of these. done clear

A) (7, 0) done clear

B) (6, 3) done clear

C) (0, 6) done clear

D) (4, 5) done clear

question_answer 4) The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let \[Z\text{ }=\text{ }px\text{ }+\text{ }qy\], where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is

A) p = q done clear

B) p = 2q done clear

C) q = 2p done clear

D) q = 3p done clear

A) The quantity in column A is greater done clear

B) The quantity in column B is greater done clear

C) The two quantities are equal done clear

D) The relationship cannot be determined on the basis of the information supplied. done clear

A) \[\left( \frac{40}{3},\frac{50}{3} \right)\] done clear

B) \[\left( \frac{50}{3},\frac{40}{3} \right)\] done clear

C) \[\left( \frac{-50}{3},\frac{40}{3} \right)\] done clear

D) \[\left( \frac{-50}{3},\frac{-40}{3} \right)\] done clear

question_answer 7) The corner points of the feasible region shown in above graph are

A) \[\left( 0,25 \right)\text{, (20,0)}\left( \frac{40}{3},\frac{50}{3} \right)\text{ }\] done clear

B) \[\left( 0,0 \right)\text{, (25,0), (0,20)}\] done clear

C) \[\left( 0,0 \right)\text{, }\left( \frac{40}{3},\frac{50}{3} \right),(0,20)\] done clear

D) \[\left( 0,0 \right)\text{, (25,0),}\left( \frac{50}{3},\frac{40}{3} \right),(0,20)\] done clear

question_answer 8) If \[Z=x+y\] be the objective function and max Z = 30. Then maximum value occurs at point

A) \[\left( \frac{50}{3},\frac{40}{3} \right)\] done clear

B) \[(0,0)\] done clear

C) (25,0) done clear

D) (0,20) done clear

question_answer 9) If \[Z=6x-9y\] be the objective function, then maximum value of Z is

A) -20 done clear

B) 150 done clear

C) 180 done clear

D) 20 done clear

question_answer 10) If Z = 6x + 3y be the objective function, then what is the minimum value of Z?

A) 120 done clear

B) 130 done clear

C) 0 done clear

D) 150 done clear

A) (6, 8) done clear

B) (5, 0) done clear

C) (0,3) done clear

D) (0,8) done clear

question_answer 12) Maximum value of Z occurs at

A) (5,0) done clear

B) (0,8) done clear

C) (0, 3) done clear

D) (6, 8) done clear

question_answer 13) Maximum of Z - Minimum of Z =

A) 58 done clear

B) 68 done clear

C) 78 done clear

D) 88 done clear

question_answer 14) The corner points of the feasible region determined by the system of linear inequalities are

A) (0, 0), (-3, 0), (3, 2), (2, 3) done clear

B) (3, 0), (3, 2), (2, 3), (0, -3) done clear

C) (0, 0), (3, 0), (3, 2), (2, 3). (0, 3) done clear

A) first and second quadrant done clear

B) first and third quadrant done clear

C) only second quadrant done clear

D) only first quadrant done clear

A) \[x+y\ge 0\] done clear

B) \[x+y<0\] done clear

C) \[x+y>0\] done clear

D) \[x+y\le 0\] done clear

A) (0, 24) done clear

B) (8, 12) done clear

C) (20, 2) done clear

question_answer 18) If the objective function of the given problem is maximise \[z=22x+18y,\] then its optimal value occur at

A) (0, 0) done clear

B) (16, 0) done clear

C) (8, 12) done clear

D) (0, 20) done clear

B) (12, 8) done clear

D) (6, 14) done clear

question_answer 20) If an LPP admits optimal solution at two consecutive vertices of a feasible region, then

A) the required optimal solution is at the mid-point of the line joining two points. done clear

B) the optimal solution occurs at every point on the line joining these two points. done clear

C) the LPP under consideration is not solvable. done clear

D) the LPP under consideration must reconstructed. done clear

A) a constant done clear

B) a function to be optimized done clear

C) a relation between the variables done clear

question_answer 22) Which of the following statement is correct

A) Every LPP has at least one optimal solution done clear

B) Every LPP has a unique optimal solution done clear

C) If an LPP has two optimal solutions, then it has infinitely many solutions done clear

question_answer 23) In solving the LPP : minimise \[f=6x+10y\] subject to contraints \[x\ge 6,\text{ }y\ge 2,\text{ }2x+y\ge 10,\text{ }x\ge 10,\,\,y\ge 0\] reducdant constraints are

A) \[x\ge 6,y\ge 2\] done clear

B) \[2x+y\ge 10,\,x\ge 0,\,\,y\ge 0\] done clear

C) \[x\ge 6\] done clear

B) (0, 8) done clear

C) (5, 0) done clear

D) (4, 10).             done clear

A) 0 done clear

B) 8 done clear

C) 12 done clear

D) 18 done clear

Study Package

Case Based (MCQs) - Linear Programming

Related question.

Reset Password.

OTP has been sent to your mobile number and is valid for one hour

Mobile Number Verified

Your mobile number is verified.

myCBSEguide

• Mathematics
• Linear Programming Class 12...

Linear Programming Class 12 Mathematics Important Questions

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Linear Programming Class 12 Mathematics Important Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in  myCBSEguide   website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 12 Linear Programming Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus  and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 12 Maths Extra Questions

Download as PDF

Class 12 Linear Programming Mathematics Extra Questions

Chapter 12 Linear Programming

• Maximum Z = 16 at (0, 4)
• Maximum Z = 19 at (1, 5)
• Maximum Z = 18 at (1, 4)
• Maximum Z = 17 at (0, 5)
• finding the optimal value (maximum or minimum) of a linear function of several variables
• finding the limiting values of a linear function of several variables
• finding the lower limit of a linear function of several variables
• finding the upper limits of a linear function of several variables
• {tex}2\frac{{10}}{{11}}{/tex}
• {tex}3\frac{{19}}{{31}}{/tex}
• {tex}3\frac{{10}}{{11}}{/tex}
• {tex}3\frac{9}{{11}}{/tex}
• 5 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 2250
• 3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950
• 6 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 2350
• 4 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 2150
• A corner point of a feasible region is a point in the region which is the ________ of two boundary lines.
• In a L.P.P, the linear inequalities or restrictions on the variables are called ________.
• In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same ________ value.
• A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximise the profit? It is being given that at least one of each must be produced.

In order to supplement daily diet, a person wishes to take some X and some wishes Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligram per tablet) are given as below:

The person needs at least 18 milligram of iron, 21 milligram of calcium and 16 milligram of vitamins. The price of each tablet of X and Y is Rs 2 and Re 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?

A factory owner purchases two types of machine A and B for his factory. The requirements and the limitations for the machines are as follows,

He has maximum area 9000 m 2 available and 72 skilled labours who can operate both the machines. How many machines of each type should he buy to maximize the daily out put?

• Minimise Z = 13x – 15y, subject to the constraints: {tex}x + y \leq 7,2 x – 3 y + 6 \geq 0 , x \geq 0 , y \geq 0{/tex} .
• A manufacturer of electronic circuits has a stock of 200 resistors, 120 transistors and 150 capacitors and is required to produce two types of circuits A and B. Type A requires 20 resistors, 10 transistors and 10 capacitors. Type B requires 10 resistors, 20 transistors and 30 capacitors. If the profit on type A circuit is Rs 50 and that on type B circuit is Rs 60, formulate this problem as a LPP so that the manufacturer can maximise his profit.
• There are two types of fertilizers F 1 and F 2 . F 1 consists of 10% nitrogen and 6% phosphoric acid and F 2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14kg of nitrogen and 14 kg of phosphoric acid for her crop. If F 1 costs Rs 6/kg and F 2 costs Rs 5/kg, determine requirements are met at a minimum cost. What is the minimum cost?

A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time an 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.  

• What number of rackets and bats must be made if the factory is t work at full capacity?
• If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
• Two tailors A and B earn Rs. 150 and Rs. 200 per day, respectively. A can stitch 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day. How many days shall each work, if it is desired to produce at least 60 shirts and 32 pants at a minimum labour cost? Make it as an LPP and solve the problem graphically.

• finding the optimal value (maximum or minimum) of a linear function of several variables Explanation: A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables.

• Intersection
• Linear constraints
• Let number of necklaces and bracelets produced by firm per day be x and y, respectively.Given that the maximum number of both necklaces and bracelets that firm can handle per day is almost 24. which means that the firm can produce maximum 24 items which includes both necklaces and bracelets per day. Hence the inequality related to the number constraint is given as {tex}\therefore{/tex} x + y {tex}\leq{/tex} 24 Also given that It takes one hour per day to make a bracelet and half an hour per day to make a necklace and maximum number of hours available per day is 16. Hence the inequality representing the hour constraint is given as {tex}\therefore{/tex} x + {tex}\frac 12{/tex} y {tex}\leq{/tex} 16 {tex}\Rightarrow{/tex} ( when multiplying throughout the inequality by 2 we get ) 2x + y {tex}\leq{/tex} 32 Also the non negative constraints which restricts the feasible region of the problem within the first quadrant is given as x {tex}\geq{/tex} 0, y {tex}\geq{/tex} 0, since the given situations are real world connected and cannot have the solution as negative which means that the values of the variables x and y are non negative. Let z be the objective function which represents the total maximum profit.Hence the equation of the profit function Z is given as {tex}\therefore{/tex} z = 100x + 300y, which is to be maximised subject to the constraints, x + y {tex}\leq{/tex} 24 2x + y {tex}\leq{/tex} 32 x, y {tex}\geq{/tex} 0

• Maximum Z = 16 at x = 4 ,y = 12
• {tex}P = 20 \times 4 + 10 \times 12{/tex} = 200

{tex}\therefore{/tex} The required linear programming problem is to minimize the wages per day. Let Z represent the objective function which represent the sum of the wages.Hence the equation of the objective function is given as (Z) = 150x + 200y Subject to constraints {tex}6 x + 10 y \geq 60{/tex} ( constraints for tailor A) ( dividing throughout by 2 we get) {tex}\Rightarrow \quad 3 x + 5 y \geq 30{/tex} {tex}4 x + 4 y \geq 32{/tex} ( constraints for tailor B) ( dividing throughout by 4 we get) {tex}\Rightarrow \quad x + y \geq 8{/tex} and {tex}x \geq 0 , y \geq 0{/tex} ( non negative constraints ,which will restrict the solution of the given inequalities in the first quadrant only) On considering the inequalities as equations, we get 3x + 5y = 50 …(i) x + y = 8 …(ii) Table for line 3x + 5y = 30 is

So, it passes through the points with coordinates (0, 6) and (10, 0). On replacing the coordinates of the origin O (0, 0) is, {tex}3 x + 5 y \geq 30{/tex} we get {tex}0 \geq 30{/tex} [which is false) So, the half plane for the inequality of the line ( i) is away from the origin, which means that the origin is not a point in the feasible region. Again, table for line ( ii) x + y = 8 is given below.

Chapter Wise Important Questions Class 12 Maths Part I and Part II

• Relations and Functions
• Inverse Trigonometric Functions
• Determinants
• Continuity and Differentiability
• Application of Derivatives
• Application of Integrals
• Differential Equations
• Vector Algebra
• Three Dimensional Geometry
• Linear Programming
• Probability

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes

Related Posts

• Chapter 12 Probability Class 12 Mathematics Important Questions
• Three Dimensional Geometry Class 12 Maths Important Questions
• CBSE Class 12 Mathematics Vector Algebra Extra Questions
• Differential Equations Class 12 Mathematics Extra Questions
• CBSE Class 12 Maths Important Questions Application of Integrals
• Integrals Class 12 Mathematics Chapter 7 Important Questions
• Class 12 Maths Application of Derivatives Extra Questions
• Continuity and Differentiability Class 12 Mathematics Extra Question

Leave a Comment

Save my name, email, and website in this browser for the next time I comment.

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Linear Programming Class 12 Maths Important Questions Chapter 12

July 7, 2022 by Sastry CBSE

Get access to Class 12 Maths Important Questions Chapter 12 Linear Programming, Linear Programming Class 12 Important Questions with Solutions Previous Year Questions will help the students to score good marks in the board examination.

Linear Programming Class 12 Important Questions with Solutions Previous Year Questions

Question 1. A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹ 100 and that on a bracelet is ₹ 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximise the profit? It is being given that at least one of each must be produced. (Delhi 2017) Answer: Let number of necklaces and bracelets produced by firm per day be x and y, respectively. Clearly, x ≥ 0, y ≥ 0 ∵ Total number of necklaces and bracelets that the firm can handle per day is atmost 24. ∴ x + y ≤ 24 Since it takes one hour to make a bracelet and half an hour to make a necklace and maximum number of hours available per day is 16. ∴ \(\frac{1}{2}\)x + y ≤ 16 ⇒ x + 2y ≤ 32 Let Z be the profit function. Then, Z = 100x + 300y ∴ The given LPP reduces to Maximise Z = 100x + 300y subject to, x + y ≤ 24 x + 2y ≤ 32 and x, y ≥ 0

Clearly, the minimum value of Z is 300 at the points (60, 0)

Question 4. Maximise and minimise Z = x + 2y subject to the constraints x + 2 y ≥ 100 2x – y ≤ 0 2x+ y ≤ 200 x, y ≥ 0 Solve the above LPP graphically. (All India 2017) Answer: Our problem is to minimise and maximise Z = x + 2y ……(i) Subject to constraints, x + 2y ≥ 100 ………..(ii) 2x – y ≤ 0 ………..(iii) 2x + y ≤ 200 ……..(iv) and x ≥ 0, y ≥ 0 ……….(v)

Table for line x + 2y = 100 is

So, the line x + 2y = 100 is passing through the points (0, 50) and (100, 0). On putting (0, 0) in the inequality x + 2y ≥ 100, we get 0 + 2 × 0 ≥ 100 ⇒ 0 ≥ 100 (which is false) So, the half plane is away from the origin. Table for line 2x – y = 0 is

So, the line 2x – y = 0 is passing through the points (0, 0) and (10, 20). On putting (5, 0) in the inequality 2x – y ≤ 0, we get 2 × 5 – 0 ≤ 0 ⇒ 10 ≤ 0 (which is false) So, the half plane is towards Y-axis. Table for line 2x + y = 200 is

The maximum value of Z is 400 at 0(0, 200) and the minimum value of Z is 100 at all the points on the line segment joining A(0, 50) and B(20, 40).

Question 5. A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 h work by a skilled man and 2 h work by a semi-skilled man. One item of model B requires 1 h by a skilled man and 3 h by a semi-skilled man. No man is expected to work more than 8 h per day. The manufacturer profit on an item of model A is t 15 and on an items of model B is ? 10. How many of items of each models should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit. (Delhi 2019) Answer: Let the company produce x items of A model and y items of B model. Maximize profit is P = 15x + 10y Now, total time spent by 5 skilled men = 2x + y and it should be less than 40. ∴ 2x + y ≤ 40 …(i) Also, the total time spent by 10 semi-skilled men = 2x + 3y and it should be less than 80. ∴ 2x + 3y ≤ 80 …(ii) Also, x ≥ 0 and y ≥ 0 Now, our problem is to maximise Z = 15x + 10y

Question 6. A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of gold while that of type B requires 1 g of silver and 2 g of gold. The company can use at the most 9 g of silver and 8 g of gold. If each unit of type A brings a profit of ₹ 40 and that of type B ₹ 50, find the number of units of each type that the company should produce to maximize profit. Formulate the above LPP and solve it graphically and also find the maximum profit. (All India 2019, CBSE 2018C) Answer: Let number of goods A = x units, and number of goods B = y units Now, the given LPP is to maximise profit: P = 40x + 50y Subject to following constraints 3x + y ≤ 9 x + 2y ≤ 8 and x ≥ 0, y ≥ 0

Clearly, P is maximum at x = 2 and y = 3 ∴ The maximum value of P is ₹ 230.

Let us find the coordinates of B which is the intersection point of 2x + y = 80 and 2x + 3y = 120. On solving these two equations, we get x = 30 and y = 20 Thus, the coordinates of B are (30, 20).

Now, let us find the value of Z at corner points, as shown in the following table.

We find that maximum value of Z is 41 at B (30, 20). Hence, the manufacturer should produce 30 packets of screw A and 20 packets of screw B to get a maximum profit of ₹ 41.

Table for line 3x + 2y = 12 or y = \(\frac{12-3 x}{2}\) is

It passes through the points (0, 6) and (4, 0). On putting (0, 0) in the inequality 3x + 2y ≤ 12, we get 0 + 0 ≤ 12 ⇒ 0 ≤ 12 [true]

So, the half plane is towards the origin. Table for line 3x + y = 9 or y = 9 – 3x is

It passes through the points (0, 9) and (3, 0). On putting (0, 0) in the inequality 3x + y ≤ 9, we get 0+ 0 ≤ 9 ⇒ 0 ≤ 9 [true]

So, the half plane is towards the origin. Also, x ≥ 0 and y ≥ 0, so the region lies in 1st quadrant. Now, on subtracting Eq. (iii) from Eq. (ii), we get (3x + 2y) – (3x + y) = 12- 9 ⇒ y = 3 Now, 3x = 12 – 2y = 12 – 2 × 3 = 6 ⇒ x = 2 Thus, the point of intersection is B(2, 3).

Hence, for maximum profit, the manufacturer should produce 2 units of product A and 3 units of product B.

Question 9. A retired person wants to invest an amount of ₹ 50000. His broker recommends investing in two types of bonds A’ and ‘B’ yielding 10% and 9% return respectively on the invested amount. He decides to invest at least ₹ 20000 in bond A’ and at least ₹ 10000 in bond ‘B’. He also wants to invest at least as much in bond A’ as in bond ‘B’. Solve this linear programming problem graphically to maximise his returns. (All Indio 2016) Answer: Let the amounts invested by the person in bonds A and B are respectively, ₹ x and ₹ y. Our problem is to maximise Z = 10% of x + 9% of y or Z = 0.1x + 0.09y Subject to constraints, x + y = 50000 x ≥ y or x – y ≥ 0 x ≥ 20000 and y ≥ 10000

Now, consider the given inequations as equations x + y = 50000 …(i) x – y = 0 …….(ii) x = 20000 …(iii) and y = 10000 …(iv)

The table for line x + y = 50000 is

∴ It passes through the points (0, 50000) and (50000, 0). The table for line x – y = 0 is

∴ It passes through the points (0, 0) and (20000, 20000). On putting (10000, 0) in the inequality x ≥ y, we get 10000 ≥ 0 (true) So, the half plane is towards the X-axis.

The line x = 20000 is perpendicular to the X-axis. On putting (0, 0) in the inequality x ≥ 20000, we cet 0 ≥ 20000 (false) So, the half plane is away from the origin.

Hence, to get a maximum returns, he has to invest ₹ 40000 in bond A and ₹ 10000 in bond B.

Let Z be the total cost of the fertilisers. Then, Z = 10x+ 8 y The LPP can be stated mathematically as Minimise Z = lOx + 8y Subject to constraints 3x + y ≥ 300, x + y ≥ 240, x ≥ 0,y ≥ 0 To solve the LPP graphically, let us convert the inequations into equations as follows : 3x + y = 300 …(i) x + y = 240 …..(ii) Table for line 3x + y = 300 is

So, it passes through (0, 300) and (100, 0). On putting (0,0) in the inequality 3x + y ≥ 300, we get 3(0) + 0 ≥ 300 ⇒ 0 ≥ 300 (which is false) So, the half plane is away from origin (1) Table for line x + y = 240 is

So, it passes through (240, 0) and (0, 240). On putting (0, 0) in x + y ≥ 240, we get 0 + 0 ≥ 240 (which is false) So, the half plane is away from origin. Also, x ≥ 0 and y ≥ 0, so the region lies in 1st quadrant.

From the table, we find that 1980 is the minimum value of Z at E (30, 210). Since, the region is unbounded, therefore, 1980 may or may not be the minimum value of Z. For this we have to check that the open half plane 10x + 8y < 1980 has any point common or not with the feasible region. Since, it has no point in common with the feasible region. So, the minimum value of Z is obtained at £(30, 210) and the miniumum value of Z is 1980. So, the farmer should use 30 kg of fertiliser A and 210 kg of fertiliser B.

Table for line 3x + y = 9 is

So, it passes through (0, 9) and (3, 0). On putting (0, 0) in 3x+ y ≥ 9, we get 0 + 0 ≥ 9 (which is false) So, the half plane is away from origin. Table for line x + y = 7 is

So, it passes through (0, 7) and (7, 0). Also, the half plane of x + y ≥ 7 is away from origin. Table for line 2x + y = 8 is

The values of the objective function at these points are given in the following table :

From the table, we find that 8 is the minimum value of Z at G(1, 6). Since, the region is unbounded, therefore, 8 may or may not be the minimum value of Z. For this we have to check, that the open half plane 2x + y < 8 has any point common or not with the feasible region.

Since, it has no point in common with the feasible region. So, Z is minimum at G(1, 6) and the minimum value of Z is 8. Hence, the person should take 1 tablet of type X and 6 tablets of type Y in order to meet the requirements at the minimum cost.

So, it passes through (0, 10/3) and (5, 0). On putting (0, 0) in the inequality 2x + 3 ≤ 10, we get 2(0) + 3(0) ≤ 10 ⇒ 0 ≤ 10 (which is true) So, the half plane is towards the origin. Table for line 3x + 2y = 10 is 10/3

So, it passes through (0, 5) and (10/3, 0) On putting (0, 0) in the inequality 3x + 2y ≤ 0, we get 3 (0) + 2 (0) = 10 ⇒ 0 ≤ 10 (which is true) So, the half plane is towards the origin. On solving Eq. (i) and Eq. (ii), we get x = 2 and y = 2 So, the intersection point is (2, 2).

Thus, profit will be maximum, when 2 packages of nuts and 2 packages of bolts are manufactured.

Question 13. Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below 2x+ 4y ≤ 8; 3x + y ≤ 6 x + y ≤ A; x ≥ 0, y ≥ 0. (Delhi 2015) Answer: We have the following LPP, Maximise, Z = 2x + 5y Subject to the constraints 2x + 4y ≤ 8 or x + 2y ≤ 4 3x + y ≤ 6 x + y ≤ 4 x ≥ 0, y ≥ 0

Now, considering the inequations as equations, we get x + 2y = 4 …(i) 3x + y = 6 …(ii) and x + y = 4 …(iii) Table for line x + 2y = 4 is

So, the line passes through (4, 0) and (0, 2). On putting (0, 0) in the inequality x + 2y ≤ 4, we get 0 + 0 ≤ 4 [which is true]

So, the half plane is towards the origin. Table for line 3x + y = 6 is

So, the line passes through (2, 0) and (0, 6). On putting (0, 0) in the inequality 3x + y ≤ 6, we get 0 + 0 ≤ 6 [which is true] So, the half plane is towards the origin. CD Table for line x + y = 4 is

On putting (0, 0) in the inequality x + y ≤ 4, we get 0 + 0 ≤ 4 [which is true] So, the half plane is towards the origin. Also, x ≥ 0, y ≥ 0, so the region lies in the 1st quadrant.

Thus, the corner points are 0(0, 0),A(2, 0), B(\(\frac{8}{5}, \frac{6}{5}\)),C(0, 2). The values of Z at corner points are as follows:

Hence, the maximum value of Z is 10.

Question 14. A company manufactures three kinds of calculators: A B and C in its two factories I and II. The company has got an order for manufacturing at least 6400 calculators of kind A 4000 of kind B and 4800 of kind C. The daily output of factory I is of 50 calculators of kind A 50 calculators of kind B and 30 calculators of kind C. The daily output of factory II is of 40 calculators of kind A 20 of kind B and 40 of kind C. The cost per day to run factory I is ₹ 12000 and of factory II is ₹ 15000. How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate this problem as an LPP and solve it graphically. All India 2015 Answer: Let factory (I) run for x days and factory (II) run fory days. Then, Total cost (in ₹) = 12000x + 15000y and our problem is to Minimise Z = 12000x + 15000y Subject to the constraints 50x + 40y ≥ 6400 or 5x + 4y ≥ 640 50x + 20y ≥ 4000 or 5x + 2y ≥ 400 30x+ 40y ≥ 4800 or 3x + 4y ≥ 480 x ≥ 0 and y ≥ 0

Now, considering the inequations as equations, we get 5x + 4y = 640 …(i) 5x + 2y = 400 …(ii) 3x + 4y = 480 …(iii)

Table for line 5x + 4y = 640 is

So, the line passes through the points (128, 0) and (0, 160). On putting (0, 0) in the inequality 5x + 4y ≥ 640, we get 0 + 0 ≥ 640 [which is false] So, the half plane is away from the origin.

Table for line 5x + 2y = 400 is

So, the line passes through the points (80, 0) and (0, 200). On putting (0, 0) in the inequality 5x + 2y ≥ 400, we get 0 + 0 ≥ 400 [which is false] So, the half plane is away from the origin.

Table for line 3x + 4y = 480 is

So, the line passes through the points (160, 0) and (0, 120). On putting (0, 0) in the inequality 3x + 4y ≥ 480, we get 0 + 0 ≥ 480 [which is false] So, the half plane is away from the origin. Also, x ≥ 0 and y ≥ 0, so the feasible region lies in the first quadrant. The point of intersection of lines (i) and (iii) is (80, 60) and lines (i) and (ii) is (32, 120).

In the table, we find that minimum value of Z is 1860000, occur at the point B(80, 60). But we can’t say that it is a minimum value of Z as region is unbounded. Therefore, we have to draw the graph of the inequality 12000x + 15000y < 1860000 or 12x + 15y < 1860

From the figure, we see that the open half plane represented by 12x + 15y < 1860 has no point in common with feasible region. Thus, the minimum value of Z is ₹ 1860000 attained at the point (80, 60). Hence, factory (I) should run for 80 days and factory (II) should run for 60 days to get a minimum cost.

Question 15. Maximise Z = 8x + 9y subject to the constraints given below: 2x + 3y ≤ 6 3x – 2y ≤ 6 y ≤ 1; x,y ≥ 0. (Foreign 2015) Answer: We have the following LPP, Maximise Z = 8x + 9y Subject to the constraints 2x + 3y ≤ 6 3x – 2y ≤ 6 y ≤1 and x, y ≥ 0

Now, considering the inequations as equations, we get 2x + 3y = 6 …………(i) 3x – 2y = 6 ………(ii) and y = 1 …(iii) Table for line 2x + 3y = 6 is

So, it passes through the points, (3, 0) and (0, 2). On putting (0,0) in the inequality 2x + 3y ≤ 6, we get 0 ≤ 6 [which is true] So, the half plane is towards the origin.

Table for line 3x – 2y = 6 is

So, it passes through the points (2,0) and (0,-3). On putting (0,0) in the inequality 3x – 2y ≤ 6, we get 0 ≤ 6 [which is true] So, the half plane is towards the origin. (1)

The line y = 1 is perpendicular to Y-axis. On putting (0,0) in the inequality y ≤ 1, we get 0 ≤ 1 [which is true] So, the half plane is towards the origin.

Also, x ≥ 0, y ≥ 0, so the feasible region lies in the first quadrant. The point of intersection of Eqs. (i) and (ii) is and Eqs.(i) \(\frac{30}{13}, \frac{6}{13}\) Eqs. (ii) and (iii) is (\(\frac{8}{3}\), 1) and Eqs. (i) and (iii) is (\(\frac{3}{2}\), 1)

The values of Z at comer points are as follows:

In the table, we find that maximum value of Z is 22.62, when x = \(\frac{30}{13}\) and y = \(\frac{6}{13}\)

Question 16. One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingredients used in making the cakes. Make it as an LPP and solve it graphically. (Delhi 2015C; All India 2014C, 2011C) Answer: We can write the given data in tabular form as follows:

Suppose the number of cakes of I kind be x and II kind be y. Then, required LPP is max (Z) = x + y

Subject to the constraints 200x + 100y ≤ 5000 ⇒ 2x + y ≤ 50 …(i) [dividing both sides by 100] 25x + 50y ≤ 1000 ⇒ x + 2y ≤ 40 …(ii) [dividing both sides by 25] and x ≥ 0, y ≥ 0

Let us consider the inequalities as equations, we get 2x + y = 50 …(iii) and x + 2y = 40 …(iv)

Table for line 2x + y = 50 is

So, it passes through the points (25, 0) and (0, 50). On putting (0,0) in the inequality 2x + y < 50, we get 0 ≤ 50 [which is true] So, the half plane is towards the origin.

Table for line x + 2y = 40 is

So, it passes through the points (40, 0) and (0,20). On putting (0,0) in the inequality x+ 2y < 40, we get 0 ≤ 40 [which is true]

The values of Z at the comer points are as follows:

From the table, maximum number of cakes is 30, which has 20 cakes of 1st kind and 10 cakes of IInd kind.

Let us consider the inequalities as equations, we get x + y = 20 …(ii) 3x + 2y = 48 …(iii) and x = 0, y = 0 …(iv)

Table for line x + y = 20 is

So, it passes through the points (0, 20) and (20, 0). On putting (0, 0) in the inequality x + y < 20, we get 0 + 0 ≤ 20 ⇒ 0 ≤ 20 [which is true] So, the half plane is towards the origin.

Table for line 3x + 2y = 48 is

From the table, maximum value of Z = ₹ 392 at point B(8,12). Hence, dealer should purchased 8 electronic and 12 manually operated sewing machines to get maximum profit.

Subject to constraints 9x + 12y ≤ 180 x + 3y ≤ 30 x ≥ 0, y ≥ 0

Let us consider the inequalities as equations, we get 9x + 12y = 180 or 3x + 4y = 60 ……….(ii) x + 3y = 30 … (iii) and x = 0,y = 0 … (iv)

Table for line 3x + 4y = 60 is

So, it passes through the points (0, 15) and (20, 0). On putting (0, 0) in the inequality 9x + 12y ≤ 180, we get 9(0) + 12(0) ≤ 180 ⇒ 0 ≤ 180 [which is true] So, the half plane is towards the origin. Table for line x + 3y = 30 is

So, it passes through the points (0,10) and (30,0). On putting (0, 0) in the inequality x + 3y ≤ 30, we get 0 + 3(0) ≤ 30 ⇒ 0 ≤ 30 [which is true] So, the half plane is towards the origin. Also, x ≥ 0 and y ≥ 0 Thus, the feasible region lies in the first quadrant. On solving Eqs. (ii) and (iii), we get x = 12 and y = 6 So, the point of intersection is (12, 6).

From graph, feasible region is OABCO, whose comer points are O(0, 0), A(20, 0), B (12, 6) and C(0,10). The values of Z at corner points are as follows:

From the table, the maximum value of Z is ₹ 1680. Hence, 12 pieces of type A and 6 pieces of type B should be manufactured per week to get a maximum profit of ₹ 1680 per week.

Let us consider the inequalities as equations, we get 2x + y = 12 …(i) 3x + 2y = 20 …(ii)

Table for 2x + y = 12 is

So, it passes through the points (0, 12) and (6, 0). On putting (0,0) in inequality 2x + y ≤ 12 we get 2(0) + 0 ≤ 12 ⇒ 0 ≤ 12 [which is true] So, the half plane is towards the origin. Table for 3x + 2y = 20 is

So, it passes through the points (0,10) and (20/3,0). On putting (0,0) in inequality 3x + 2y < 20, we get 3 (0) + 2 (0) ≤ 20⇒ 0 ≤ 20 [which is true] So, the half plane is towards the origin. Also, x ≥ 0, y ≥ 0, so the feasible region lies in first quadrant. On solving Eqs. (i) and (ii), we get x = 4 and y = 4 So, the intersection point is (4, 4)

The values of Z at the corner points are as follows:

From table, maximum value of Z is 160 at B(4, 4). Hence, the manufacturer should produce 4 pedestal lamps and 4 wooden shades daily to get maximum profit.

Question 20. A decorative item dealer deals in two items A and B. He has ₹ 15000 to invest and a space to store at the most 80 pieces. Item A costs him ₹ 300 and item B costs him ₹ 150. He can sell items A and B at respective, profits of ₹ 50 and ₹ 28. Assuming he can sell all he buys, formulate the linear programming problem in order to maximise his profit and solve it graphically. (Delhi 2012C) Answer: Let the number of items of the type A and B be x and y, respectively. Then the required LPP is Maximise Z = 50x + 28y Subject to the constraints, x + y ≤ 80, 300x + 150y ≤ 15000 or 2x + y ≤ 100, x ≥ 0, y ≥ 0 Let us consider the inequalities as equations, we get x + y = 80 …………..(i) 2x + y = 100 …………..(ii) Table for line x + y = 80 is

So, it passes through the points (0, 80) and (80, 0). On putting (0, 0) in the inequality x + y ≤ 80, we get 0+ 0 ≤ 80 ⇒ 0 ≤ 80 (which is true) So, the half plane is towards the origin.

Table for line 2x + y = 100

From the above table, the maximum value of Z is 2680 at P(20, 60), i.e. when 20 items of type A and 60 items of type B are purchased and sold. He get maximum profit.

So, it passes through (0, 10/3) and (5, 0). On putting (0, 0) in the inequality 2x + 3 ≤ 10, we get 2(0) + 3(0) ≤ 10 ⇒ 0 ≤ 10 (which is true) So, the half plane is towards the origin. (1) Table for line 3x + 2y = 10 is 10/3

Thus, profit will be maximum, when 2 packages of nuts and 2 packages of bolts are manufactured. Hint: Let the manufacturer produces x packages of nuts and y packages of bolts. The required LPP is Maximize Z = 17.50x + 7y Subject to constraints x + 3y ≤ 12 3x + y ≤ 12 and x,y ≥ 0 [The manufacturer should produce 3 packages of nuts and 3 packages of bolts each day for maximum profit = ₹ 73.50]

Thus, profit will be maximum, when 2 packages of nuts and 2 packages of bolts are manufactured. Hint: Let the manufacturer produces x packages of nuts and y packages of bolts. The required LPP is Maximize Z = 2.5x + y Subject to constraints x + 3y ≤ 12 3x + y ≤ 12 and x, y ≥ 0 [The manufacturer should produce 3 packages of nuts and 3 packages of bolts each day for maximum profit = ₹ 10.5]

So, it passes through the points (0, 8) and (4, 0). On putting (0,0) in the inequality 2x + y ≥ 8, we get 0 ≥ 8 [which is false] So, the half plane is away from the origin.

Table for line x + 2y = 10 is

So, it passes through the points (10, 0) and (0, 5). On putting (0,0) in the inequality x + 2y ≥ 10, we get 0 ≥ 10 [which is false] So, the half plane is away front the origin. Also, x, y ≥ 0, so the feasible region lies in the first quadrant. On solving Eq. (i) and Eq. (ii), we get x = 2 and y = 4 So, these lines intersect at P (2, 4).

From table, the minimum value of Z is 38. As the feasible region is unbounded, therefore 38 may or may not be the minimum value of Z. For this, we draw a dotted graph of the inequality 5x + 7y < 38 and check whether the resulting half plane has point in common with the feasible region or not. It can be seen that the feasible region has no common point with 5x + 7y < 38. Hence, the minimum cost is ₹ 38, when x = 2 and y = 4.

Hint: Suppose the diet contains x kg of food I and y kg of food II. Then, the required LPP is minimise (Z) = 50x + 70y Subject to contraints 2x+ y ≥ 8, x+ 2y ≥ 10 and x ≥ 0,y ≥ 0 [The minimum cost of the mixture is X 380, when the mixture contains 2 kg of food I and 4 kg of food II]

Hint: Let x be the number of tennis rackets and y be the cricket bats produced in one day in the factory. The required LPP is maximise (Z) = 20x + 10y Subject to constraints 1.5x + 3y ≤ 42 3x + y ≤ 24 and x ≥ 0, y ≥ 0 [For maximum profit of X 200, 4 tennis rackets and 12 cricket bats must be produced.]

Table for line x + y = 250 is

So, this line passes through the points (0, 250) and (250, 0). On putting (0,0) in the inequality x + y ≤ 250, we get 0 ≤ 250 [which is true] So, the half plane is towards the origin. (1) Table for line 5x + 8y = 1400 is

So, this line passes through the points (280, 0) and (0, 175). On putting (0,0) in 5x + 8y ≤ 1400, we get ⇒ 0 ≤ 1400 [which is true] So, the half plane is towards the origin. Also, x, y ≥ 0, so the feasible region lies in the first quadrant. On solving Eqs. (ii) and (iii), we get x = 200 and y = 50 So, the intersection point is E(200, 50).

From the table, maximum value of Z is 1150000 at E (200,50). Hence, the profit is maximum, i.e. ₹ 1150000, when 200 desktop computers and 50 portable computers are stocked.

Question 27. A dealer deals in two items A and B. He has ₹ 15000 to invest and a space to store atmost 80 pieces. Item A costs him ₹ 300 and item B costs him ₹ 150. He can sell items A and B at profits of ₹ 40 and ₹ 25, respectively. Assuming that he can sell all that he buys, formulate the above as a linear programming problem for maximum profit and solve it graphically. (Delhi 2010C) Answer: He sell 20 items of type A and 60 items of type B. Maximum profit = ₹ 2300

Question 28. A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is atmost 24. It takes 1 h to make a ring and 30 min to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is ₹ 300 and that on a chain is ₹ 190, then find the number of rings and chains that should be manufactured per day so as to earn the maximum profit. Make it as an LPP and solve it graphically. (Delhi 2010) Answer: Let the firm manufactures x gold rings and y chains per day. The given data can be written in tabular form as follows:

The required LPP is maximise, Z = 300x + 190y Subject to constraints x + y ≤ 24 x + -y ≤ 16 or 2x + y ≤ 32 and x ≥ 0, y ≥ 0 On considering the inequalities as equations. we get x + y = 24 . ..(i) and 2x + y = 32 ………….(ii)

Table for line x + y = 24 is

So, it passes through (0, 24) and (24, 0). On putting (0, 0) in the inequality x + y ≤ 24, we get 0 ≤ 24 (which is true) So, the half plane is towards the origin, (1)

Again, table for line 2x + y = 32 is

So, it passes through (0, 32) and (16, 0).

On putting (0, 0) in the inequality 2x + y ≤ 32 we get 0 ≤ 32 (which is true) So, the half plane is towards the origin. On solving Eq. (i) and Eq. (ii), we get x = 8 andy = 16 So, the point of intersection is B(8, 16)

From the table, maximum value of z is 5440 at B(8, 16). Hence, The manufacturer earns the maximum profit ₹ 5440, when he manufactures 8 gold rings and 16 chains per day.

Question 29. A library has to accommodate two different types of books on a shelf. The books are 6 cm and 4 cm thick and weight 1 kg and 1 \(\frac{1}{2}\) kg each, respectively. The shelf is 96 cm long and atmost can support a weight of 21 kg. How should the shelf be filled with the hooks of two types in order to include the greatest number of books? Make it as an LPP and solve it graphically. (All India 2010C) Answer: Let number of books of two types be x and y, respectively. The required LPP is maximise Z = x + y Subject to constraints 6x + 4y ≤ 96 or 3x + 2y ≤ 48 x + – y ≤ 21 or 2x + 3y ≤ 42 2 and x, y ≥ 0 On considering the inequalities as equations, we get 3x + 2y = 48 …(i) 2x + 3y = 42 …(ii) Table for line 3x + 2y = 48 is

So, it passes through (0, 24) and (16, 0). On putting (0, 0) in 3x + 2y ≤ 48, we get 0 + 0 ≤ 48 ⇒ 0 ≤ 48 (which is true) So, the half plane is towards the origin.

Table for 2x + 3y = 42 is

From the table, the maximum value of Z is 18 at B(12, 6). Hence, The maximum number of books is 18 and number of books of I type is 12 and books of II type is 6.

Question 30. One kind of cake requires 300 g of flour and 15 g of fat, another kind of cake requires 150 g of flour and 30 g of fat. Find the maximum number of cakes which can be made from 7.5 kg of flour and 600 g of fat, assuming that there is no shortage of other ingredients used in making the cakes. Make it as an LPP and solve it graphically. (All India 2010) Answer: The maximum number of cakes is 30 in which number of cake of first kind of cakes is 20 and second kind is 10.

Free Resources

NCERT Solutions

Quick Resources

Talk to our experts

1800-120-456-456

Important Questions for CBSE Class 12 Maths Chapter 12 - Linear Programming 2024-25

• Class 12 Important Question
• Chapter 12: Linear Programming

CBSE Class 12 Maths Chapter-12 Important Questions - Free PDF Download

Download free PDF of Important Questions for CBSE Class 12 Maths Chapter 12 - Linear Programming prepared by expert Maths teachers from the latest edition of CBSE (NCERT) books . Register online for Maths tuition on Vedantu.com to score more marks in CBSE board examination .

Class 12 Maths Chapter 12 Linear Programming is an important chapter that students should provide attention to. This chapter develops the fundamental concepts related to linear programming that will be very useful in the future. To make this chapter easier to prepare, download and solve the CBSE Class 12 Maths Chapter 12 Linear programming Important Questions developed by the experts. Refer to the solutions to learn the answering methods and score more in the exams.

Download CBSE Class 12 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 12 Maths Important Questions for other chapters:

Related Chapters

Study Important Questions for class 12 Mathematics Chapter 12 – Linear Programming

Long Answer Type Questions (6 Marks)

1. Solve the following L.P.P. graphically

Minimise and maximise $\quad {\text{z}} = 3x + 9y$

Subject to the constraints $\quad x + 3y \leqslant 60$

                                                   $x + y \geqslant 10$

                                                   $x \leqslant y$

                                                   $x \geqslant 0,y \geqslant 0$

Maximize $Z = 3x + 9y$

Minimize $Z = 3x + 9y$

Subject to,

$\begin{array}{*{20}{c}} {x + 3y \leqslant 60} \\ {x + y \geqslant 10} \\ {x \leqslant y} \\ {x \geqslant 0,\quad y \geqslant 0} \end{array}$

\[x{\text{ }} + {\text{ }}3y{\text{ }} \leqslant {\text{ }}60\]

\[x{\text{ }} + {\text{ }}y{\text{ }} \geqslant {\text{ }}10\]

\[x{\text{ }} \leqslant {\text{ }}y\]

$\therefore {\text{Z}} = 60$ is minimum at $(5,5)$

$Z$= 60 is maximum at two points $(15,15){\text{ \&  }}(0,20)$

$\therefore {\text{Z}} = 180$ is maximum at all points joining $(15,15){\text{ \&  }}(0,20)$.

2. Determine graphically the minimum value of the objective function z = – 50x + 20 y, subject to the constraints

Minimum $Z =  - 50x + 20y$

$\begin{array}{*{20}{l}} {2x - y \geqslant - 5} \\ {3x + y \geqslant 3} \\ {2x - 3y \leqslant 12} \\ {x \geqslant 0,y \geqslant 0} \end{array}$

\[2x{\text{ }}--{\text{ }}y \geqslant  - 5\]

\[3x{\text{ }}--{\text{ }}3y \geqslant 3\]

\[2x{\text{ }}--{\text{ }}3y \leqslant 12\]

(Image will be uploaded soon)

As the region that is feasible is unbounded.

Hence $ - 300$ may or may not be the minimum value of $Z$.

To check this, we graph inequality.

$ - 50x + 20y <  - 300$

As, there are some common points between the feasible region 

the inequality.

Hence, \[300\]is not the minimum value of $Z$.

$\therefore {\mathbf{Z}}$ has no minimum value.

3. Two tailors A and B earn \[{\mathbf{Rs}}.{\text{ }}{\mathbf{150}}{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{Rs}}.{\text{ }}{\mathbf{200}}\] per day respectively. A can stitch 6 shirts and 4 pants per day, while B can stitch \[{\mathbf{10}}{\text{ }}{\mathbf{shirts}}{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{4}}{\text{ }}{\mathbf{pants}}\]per day. Formulate the above L.P.P. mathematically and hence solve it to minimize the labour cost to produce at least \[{\mathbf{60}}{\text{ }}{\mathbf{shirts}}{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{32}}{\text{ }}{\mathbf{pants}}\].

Let Tailor A work for x days

Tailor ${\text{B}}$ work for y days

Two tailors $A$ and $B$ earn \[150{\text{ }}and{\text{ }}200\]per day respectively.

And, we need to minimize labour cost

So, our equation will be

$\therefore {\text{Z}} = 150{\text{x}} + 200{\text{y}}$

Now, according to question

A can stitch 6 shirts and 4 pants per day, while $B$ can stitch 10 shirts

and 4 pants per day. We need to minimize the labour cost to produce

at least 60 shirts and 32 pants and solve it graphically

Shirts                         Pants

Min Shirts = 60        Min Pants = 32

\[\begin{array}{*{20}{l}} {6x{\text{ }} + {\text{ }}10y \geqslant 60\;\;\;\;\;\;\;\;\;\;\;\;\;4x{\text{ }} + {\text{ }}4y \geqslant 32} \\ {3x{\text{ }} + {\text{ }}5y \geqslant 30\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x{\text{ }} + {\text{ }}y \geqslant 8} \end{array}\]

\[x \geqslant 0,{\text{ }}y \geqslant 0\]Combining all constraints:

$\operatorname{Min} Z = 150x + 200y$

Subject to constraints,

\[\begin{array}{*{20}{l}} {3x{\text{ }} + {\text{ }}5y \geqslant 30} \\ {x{\text{ }} + {\text{ }}y \geqslant 8} \\ {\& {\text{ }}x \geqslant {\text{ }}0,{\text{ }}y{\text{ }} \geqslant {\text{ }}0} \end{array}\]

\[3x{\text{ }} + {\text{ }}5y \geqslant 30\]

\[x{\text{ }} + {\text{ }}y \geqslant 8\]

Hence \[1350\]may or may not be the minimum value of $Z$.

$\begin{array}{*{20}{l}} {150x + 200y \leqslant 1350} \\ {{\text{ i}}{\text{.e}}{\text{. }}3x + 4y \leqslant 27} \\ {3x + 4y \leqslant 27} \end{array}$

\[3x{\text{ }} + {\text{ }}4y \leqslant 27\]

As there is no common point between the feasible region & the

inequality.

Hence, 1350 is the minimum value of $Z$.

Hence, cost will be minimum if

Number of days tailor $A$ works for $ = 5$

Number of days tailor $B$ works for $ = 3$

Minimum Cost = \[Rs{\text{ }}1350\] at 5 days of A and 3 days of B.

4. There are two types of fertilizer’s A and B. A consists of 10% nitrogen and 6% phosphoric acid and B consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If A costs Rs. 6 kg and B costs Rs. 5 kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at minimum cost. What is the minimum cost?

Let the fertilizer ${{\text{F}}_1}$ requirement be x kg

fertilizer ${{\text{F}}_2}$ requirement be y kg

According To questions:

Also, \[x{\text{ }} \geqslant {\text{ }}0,{\text{ }}y \geqslant {\text{ }}0\]

As we need to Minimize the cost,

hence the function used here is Minimize $z$.

Cost of fertilizer ${{\text{F}}_1} = $ Rs 6 kg

Cost of fertilizer ${{\text{F}}_2} = $ Rs 5 kg

Minimize $z = 6x + 5y$

Combining all Constraints:

$\operatorname{Min} Z = 6x + 5y$

Subject to constraints

$\begin{array}{*{20}{l}} {2x + y \geqslant 280} \\ {3x + 5y \geqslant 700} \\ {x,y \geqslant 0} \end{array}$

\[2x{\text{ }} + {\text{ }}y \geqslant 280\]

\[3x{\text{ }} + {\text{ }}5y \geqslant 700\]

As the feasible area is unbounded

Hence 1000 may or may not be the minimum value of $Z$.

For this, we need to graph inequality

Since, there is no common point between the feasible region and

$6x + 5y < 1000$

Hence the cost will be minimum, if

Fertilizer ${{\text{F}}_1}$ used $ = 100\;{\text{kg}}$

Fertilizer ${{\text{F}}_2}$ used 8= 80kg

Minimum Cost = Rs 1000.

5. A man has Rs. 1500 to purchase two types of shares of two different companies S1 and S2. Market price of one share of S1 is Rs 180 and S2 is Rs. 120. He wishes to purchase a maximum of ten shares only. If one share of type S1 gives a yield of Rs. 11 and of type S2 yields Rs. 8 then how much shares of each type must be purchased to get maximum profit? And what will be the maximum profit?

Let the S1 purchased be x and S2 purchased be y 

Then, \[180x{\text{ }} + {\text{ }}120y <  = {\text{ }}1500\]

Also, \[x{\text{ }} + {\text{ }}y{\text{ }} <  = 10\]

 \[x >  = 0{\text{ }}and{\text{ }}y >  = 0\]

\[Profit{\text{ }} = {\text{ }}11x{\text{ }} + {\text{ }}8y\]

 Now, if you will draw the graph using the 4 constraint equations mentioned above, 

the common region would be a closed surface with A (0,0), B (8.33,0), C (5,5), D (0,10)

Hence, the profit would be maximum at one of the vertices only

\[\begin{array}{*{20}{l}} {Profit{\text{ }}at{\text{ }}A{\text{ }}\left( {0,0} \right){\text{ }} = {\text{ }}11\left( 0 \right){\text{ }} + {\text{ }}8\left( 0 \right){\text{ }} = {\text{ }}0} \\ {Profit{\text{ }}at{\text{ }}B{\text{ }}\left( {8.33,0} \right){\text{ }} = {\text{ }}11\left( {8.33} \right){\text{ }} + {\text{ }}8\left( 0 \right){\text{ }} = {\text{ }}91.63} \\ {Profit{\text{ }}at{\text{ }}C{\text{ }}\left( {5,5} \right){\text{ }} = {\text{ }}11\left( 5 \right){\text{ }} + {\text{ }}8\left( 5 \right){\text{ }} = {\text{ }}95} \\ {Profit{\text{ }}at{\text{ }}D{\text{ }}\left( {0,10} \right){\text{ }} = {\text{ }}11\left( 0 \right){\text{ }} + {\text{ }}8\left( {10} \right){\text{ }} = 80} \\ \; \end{array}\]

Hence, the maximum profit is Rs 95 at point C and hence, the man should purchase 5 Share of S1 and S2 to maximise his profit.

6. A company manufactures two types of lamps say A and B. Both lamps go through a cutter and then a finisher. Lamp A requires 2 hours of the cutter’s time and 1 hours of the finisher’s time. Lamp B requires 1 hour of cutter’s and 2 hours of finisher’s time. The cutter has 100 hours and finishers has 80 hours of time available each month. Profit on one lamp A is Rs. 7.00 and on one lamp B is Rs. 13.00. Assuming that he can sell all that he produces, how many of each type of lamps should be manufactured to obtain maximum profit?

Let the company manufacture $x$ cutters of type $A$ and $y$ finishers of type $B$.

$x \geqslant 0$ and $y \geqslant 0$

The given information can be complied in a table as follows.

The profit on type A cutters is Rs 5 and on type B finishers is Rs 6. Therefore, the

constraints are

$2x + y \leqslant 100$

$x + 2y \leqslant 80$

Total profit, $z = 7x + 13y$

The mathematical formulation of the given problem is Maximize $z = 7x + 13y \ldots (1)$

subject to the constraints, $2x + y \leqslant 100$... (2)

$x + 2y \leqslant {80_ \ldots }$

$x,y \geqslant 0 \ldots $ (4)

The feasible region determined by the system of constraints is as follows.

As per the constraints the feasible solution is the shaded region. Possible points for maximizing ${\text{Z}}$ are $A(0,40),B(40,20),C(50,0)$

$\begin{array}{*{20}{l}} {{\text{Z}}{]_A} = 7 \times 0 + 13 \times 40 = 520} \\ {{\text{Z}}{]_B} = 7 \times 40 + 13 \times 20 = 540} \\ {{\text{Z}}{]_C} = 7 \times 50 + 13 \times 0 = 350} \end{array}$

Hence profit is maximum of Rs 540 when the dealer purchases 40 Lamp A and 20 Lamp B.

7. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 5760 to invest and has space for almost 20 items. A fan and sewing machine cost Rs. 360 and Rs. 240 respectively. He can sell a fan at a profit of Rs. 22 and sewing machine at a profit of Rs. 18. Assuming that he can sell whatever he buys, how should he invest his money to maximize his profit?

Let the no. of fans purchased by the dealer $ = {\text{x}}$ 

and number of sewing machines purchased $ = {\text{y}}$ 

then the L.P.P. is formulated as 

$Z = 22x + 18y$ to be maximised subject to constraints

  $\begin{array}{*{20}{l}} {x + y \leqslant 20} \\ {360x + 240y \leqslant 5760} \end{array}$ $...(i)$ (space only for 20 items)

$ \Rightarrow \quad 3x + 2y \leqslant 48$        $ \ldots (ii)$

$x \geqslant 0,y \geqslant 0$                   $ \ldots (iii)$

We plot the graph of the constraints.

As per the constraints the feasible solution is the shaded region. Possible points for maximising $Z$ are $A(0,20),B(8,12),C(16,0)$

$\begin{array}{*{20}{l}} {{\text{Z}}{]_A} = 22 \times 0 + 18 \times 20 = 360} \\ {{\text{Z}}{]_B} = 22 \times 8 + 18 \times 12 = 392} \\ {{\text{Z}}{]_C} = 22 \times 16 + 18 \times 0 = 352} \end{array}$

Hence profit is maximum of Rs 392 when the dealer purchases 8 fans and 12 sewing machines.

8. If a young man rides his motorcycle at 25 km/h, he has to spend Rs. 2 per km on petrol. If he rides at a faster speed of 40 km/h, the petrol cost increases to Rs. 5 per km. He has Rs. 100 to spend on petrol and wishes to find the maximum distance he can travel within one hour. Express this as L.P.P. and then solve it graphically.

Let young man drives $x\;{\text{km}}$ at a speed of 25 km/ hr. and y km at a speed of 40km/hr.

$x,y \geqslant 0$

It is given that, he spends Rs 2 per km if he drives at a speed of $25\;{\text{km}}/{\text{hr}}$ and Rs 5 per km if he drives at a speed of $40\;{\text{km}}/{\text{hr}}$. Therefore, money spent by him when he travelled $x\;{\text{km}}$ and $y\;{\text{km}}$ are Rs 2x and Rs 5y respectively.

It is given that he has a maximum of Rs 100 to spend.

Thus, $2x + 5y \leqslant 100$

Time spent by him when travelling with a speed of $25\;{\text{km}}/{\text{hr}} = \dfrac{x}{{25}}{\text{hr}}$

Time spent by him when travelling with a speed of $40\;{\text{km}}/{\text{hr}} = \dfrac{y}{{40}}{\text{hr}}$

Also, the available time is 1 hour.

$\dfrac{x}{{25}} + \dfrac{y}{{40}} \leqslant 1$

Or, $40x + 25y \leqslant 1000$

The distance covered is $Z = x + y$ which is to be maximized.

Thus, the mathematical formulation of the given linear programming problem is $\operatorname{Max} Z = x + y$ subject to

$2x + 5y \leqslant 100$

$40x + 25y \leqslant 1000$

First, we will convert inequations as follows:

$2x + 5y = 100$

$40x + 25y = 1000$

$x = 0$ and $y = 0$

The region represented by $2x + 5y \leqslant 100$

The line $2x + 5y = 100$ meets the coordinate axes at $A(50,0)$ and $B(0,20)$ respectively. By joining these points, we obtain the line $2x + 5y = 100$. Clearly $(0,0)$ satisfies the $2x + 5y = 100.$ So, the region which contains the origin represents the solution set of the inequation $2x + 5y \leqslant 100$

The region represented by $40x + 25y \leqslant 1000$

The line $40x + 25y = 1000$ meets the coordinate axes at $C(25,0)$ and $D(0,40)$ respectively. By joining these points, we obtain the line $2x + y = 12$. Clearly $(0,0)$ satisfies the $40x + 25y = 1000$. So, the region which contains the origin represents the solution set of the inequation $40x + 25y \leqslant 1000$

The region represented by $x \geqslant 0,y \geqslant 0$ :

Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations $x \geqslant 0$ and $y \geqslant 0$.

The feasible region determined by the system of constraints

$2x + 5y \leqslant 100,40x + 25y \leqslant 1000,x \geqslant 0$ and $y \geqslant 0$ are as follows

The corner points are O (0, 0), B (0, 20), E (50 3, 40 3), (503,403), and C (25, 0). The value of Z at these corner points are as follows:

The maximum value of $Z$ is 30 which is attained at $E$.

Thus, the maximum distance travelled by the young man is $30{\text{kms}}$, if he drives $\dfrac{{50}}{3}$ ${\text{km}}$ at a speed of $25\;{\text{km}}/{\text{hr}}$ and $\dfrac{{40}}{3}\;{\text{km}}$ at a speed of $40\;{\text{km}}/{\text{hr}}$.

9. A producer has 20 and 10 units of labour and capital respectively which he can use to produce two kinds of goods X and Y. To produce one unit of X, 2 units of capital and 1 unit of labour is required. To produce one unit of Y, 3 units of labour and one unit of capital is required. If X and Y are priced at Rs. 80 and Rs. 100 per unit respectively, how should the producer use his resources to maximize the total revenue?

Let ${x_1}$ and ${y_1}$ units of goods $x$ and $y$ were produced respectively. Number of units of goods cannot be negative.

Therefore, 

${x_1},{y_1} \geqslant 0$

To produce one unit of x, 2 units of labour and for one unit of y, 1 units of labour are required $2{x_1} + {y_1} \leqslant 20$ To produce one unit of x, 3 units of capital is required and 1 unit of capital is required to produce one unit of $y$ $3{x_1} + {y_1} \leqslant 10$

If $x$ and $y$ are priced at Rs 80 and Rs 100 per unit respectively, 

Therefore, cost of ${x_1}$ and ${y_1}$ units of goods $x$ and $y$ is Rs $80{x_1}$ and Rs $100{y_1}$respectively.

Total revenue $ = Z = 80{x_1} + 100{y_1}$ which is to be maximized.

Thus, the mathematical formulation of the given linear programming problem is $\operatorname{Max} {\text{Z}} = 80{x_1} + 100{y_1}$

subject to $2{x_1} + {y_1} \leqslant 20$

$3{x_1} + {y_1} \leqslant 10$

First, we will convert inequations into equations as follows:

$2{x_1} + {y_1} = 20,3{x_1} + {y_1} = 10,x = 0$ and $y = 0$

Region represented by $2{x_1} + {y_1} \leqslant 20$ :

The line $2{x_1} + {y_1} = 20$ meets the coordinate axes at $A(10,0)$ and $B(0,20)$ respectively. By joining these points, we obtain the line

$2{x_1} + {y_1} = 20.$

Clearly $(0,0)$ satisfies the $2{x_1} + {y_1} = 20.$ So, the region which contains the origin represents the solution set of the inequation $2{x_1} + {y_1} \leqslant 20$

Region represented by $3{x_1} + {y_1} \leqslant 10$ :

The line $3{x_1} + {y_1} = 10$ meets the coordinate axes at

$C\left( {\dfrac{{10}}{3},0} \right)$ and $D(0,10)$ respectively.

 By joining these points, we obtain the line $3{x_1} + {y_1} = 10.$ 

Clearly $(0,0)$ satisfies the inequation $3{x_1} + {y_1} \leqslant 10.$ 

So, the region which contains the origin represents the solution

set of the inequation $3{x_1} + {y_1} \leqslant 10$

Region represented by ${x_1} \geqslant 0$ and ${y_1} \geqslant 0$ :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations $x \geqslant 0$, and $y \geqslant 0$.

The feasible region determined by the system of constraints $2{x_1} + {y_1} \leqslant 20,3{x_1} + {y_1} \leqslant 10,x \geqslant 0$ and $y \geqslant 0$ are as follows.

The maximum value of $Z$ is \[760\]which is attained at $E(2,6)$.

Thus, the maximum revenue is \[Rs{\text{ }}760\]obtained when 2 units of $x$ and 6 units of $y$.

10. A factory owner purchases two types of machines A and B for his factory. The requirements and limitations for the machines are as follows:

He has maximum area of 9000 m2 available and 72 skilled laborer’s who can operate both the machines. How many machines of each type should he buy to maximize the daily output?

Ans: Let the owner buys x machines of type A and y machines of type B.

$\begin{array}{*{20}{c}} {1000x + 1200y}&{ \leqslant 9000}&{ \ldots (i)} \\ {12x + 8y}&{ \leqslant 72}&{ \ldots (ii)} \end{array}$

Objective function is to be maximize $z = 60x + 40y$ From $(i)$

$10x + 12y \leqslant 90$

$\begin{array}{*{20}{l}} {5x + 6y \leqslant 45} \\ {3x + 2y \leqslant 18} \end{array}$       $...(iii)$[ from $(ii)]$

 We plot the graph of inequations shaded region in the feasible solutions (iii) and (iv).

The shaded region in the figure represents the feasible region which is bounded. 

Let us now evaluate ${\text{Z}}$ at each corner point.

at $(0,0)Z$ is $60 \times 0 + 40 \times 0 = 0$

${\text{Z}}$ at $\left( {0,\dfrac{{15}}{2}} \right)$ is $60 \times 0 + 40 \times \dfrac{{15}}{2} = 300$

${\text{Z}}$ at $(6,0)$ is $60 \times 6 + 40 \times 0 = 360$

${\text{Z}}$ at $\left( {\dfrac{9}{4},\dfrac{{45}}{8}} \right)$ is $60 \times \dfrac{9}{4} + 40 \times \dfrac{{45}}{8} = 135 + 225 = 360$

$ \Rightarrow \max .{\text{Z}} = 360$

Therefore, there must be either x=6, Y=0 or x=9/4, y=45/8 but second case is not possible as x and y are whole numbers. 

Hence there must be 6 machines of type A and no machine of type B is required for maximum daily output.

11. A manufacturer makes two types of cups A and B. Three machines are required to manufacture the cups and the time in minutes required by each in as given below:

Each machine is available for a maximum period of 6 hours per day. If the profit on each cup A is 75 paise and on B is 50 paise, find how many cups of each type should be manufactured to maximize the profit per day.

Let number of tea-cups A produced Be $x$ and number of tea-cups B produced be y

$\therefore $ It is to maximize $\quad {\text{Z}} = 1.50x + 1.y$

i.e., $\quad \operatorname{Max} {\text{Z}} = 1.50x + y$

Subject to $\quad 12x + 6y \leqslant 360$

                   $18x + 0.y \leqslant 360$

And            $\quad 6x + 9y \leqslant 360$

where        $\quad x,y \geqslant 0$

$\therefore $ Given problems can be arranged $\operatorname{Max} Z = 1.50x + y$

Subject to    $\quad 2x + y \leqslant 60$

                      $x \leqslant 20$

And               $\quad 2x + 3y \leqslant 120$

                       $x \geqslant 0;y \geqslant 0$

Consider the equation $2x + y = 60$

when $\quad x = 0\quad $ then $y = 60$

when $\quad y = 0\quad $ then $\quad x = 30$

Table of solutions

$2x + 3y = 120$

when $x = 0$ then $y = 40$

when $y = 0$ then $x = 60$

The corner points of the feasible region from the graph are $O(0,0),A(20,0),B(20,20)$, $C(15,30)$ and $D(0,40)$

At $O(0,0);Z = 1.5(0) + 0 = 0$

At ${\text{A}}(20,0);{\text{Z}} = 1.5(20) + 0 = 30$

At $B(20,20);Z = 1.5(20) + 20 = 50$

At $C(15,30);Z = 1.5(15) + 30 = 52.50$

At $D(0,40);Z = 1.5(0) + 40 = 40$

Maximum Profit $(Z) = $ Rs.52.50 at $C(15,30)$

Cup A: 15; Cup B: 30

12. A company produces two types of belts A and B. Profits on these belts are Rs. 2 and Rs. 1.50 per belt respectively. A belt of type A requires twice as much time as belt of type B. The company can produce almost 1000 belts of type B per day. Material for 800 belts per day is available. Almost 400 buckles for belts of type A and 700 for type B are available per day. How much belts of each type should the company produce so as to maximize the profit?

Let the company produces $x$ belts of types $A$ and $y$ belts of type B. Number of belts cannot be negative. 

Therefore, $x,y \geqslant 0$.

It is given that leather is sufficient only for 800 belts per day (both $A$ and $B$ combined).

$x + y \leqslant 800$

It is given that the rate of production of belts of type $B$ is 1000 per day. Hence the time taken to produce $y$ belts of type $B$ is $\dfrac{y}{{1000}}$

And, since each belt of type $A$ requires twice as much time as a belt of type $B$, the rate of production of belts of type $A$ is 500 per day and therefore, total time taken to produce $x$ belts of type $A$ is $\dfrac{x}{{500}}$

Thus, we have,

$\dfrac{x}{{500}} + \dfrac{y}{{1000}} \leqslant 1$

Or, $2x + y \leqslant 1000$

Belt A requires fancy buckle and only 400 fancy buckles are available for this per day.

$x \leqslant 400$

For Belt of type B only 700 buckles are available per day.

$y \leqslant 700$

profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively.

Therefore, profit gained on $x$ belts of type $A$ and $y$ belts of type $B$ is Rs 2x and Rs 1.50 y respectively. 

Hence, the total profit would be $\operatorname{Rs} (2x + 1.50y)$. Let $Z$ denote the total profit

$z = 2x + 1.50y$

Thus, the mathematical formulation of the given linear programming problem is;

$\operatorname{Max} {\text{Z}} = 2{\text{x}} + 1.50{\text{y}}$ subject to

$2x + y \leqslant 1000$

First, we will convert these inequations into equations as follows:

$x + y = 800$

$2x + y = 1000$

Region represented by $x + y = 800$

The line $x + y = 800$ meets the coordinate axes at $A(800,0)$ and $B(0,800)$ respectively. By joining these points, we obtain the line $x + y = 800$. Clearly $(0,0)$ satisfies the $x + y \geqslant 800$.

 So, the region which contains the origin represents the solution set of the inequation $x + y \geqslant 800$.

Region represented by $2x + y \geqslant 1000$

The line $2x + y = 1000$ meets the coordinate axes at $C(500,0)$ and $D(0,1000)$ respectively. 

By joining these points, we obtain the line $2x + y = 1000$. Clearly $(0,0)$ satisfies the $2x + y \geqslant 1000$. So, the region which contains the origin represents the solution set of the inequation $2x + y \geqslant 1000$.

Region represented by $x \leqslant 400$

The line $x = 400$ will pass through $(400,0)$. The region to the left of the line $x = 400$ will satisfy the inequation $x \leqslant 400$

Region represented by $y \leqslant 700$

The line $y = 700$ will pass through $(0,700)$. The region to the left of the line $y = 700$

will satisfy the inequation $y \leqslant 700$

Region represented by $x \geqslant 0,y \geqslant 0$ :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations $x \geqslant 0$ and $y \geqslant 0$.

The feasible region determined by the system of constraints $x + y \leqslant 800,2x + y \leqslant $ $1000,x \leqslant 400,y \leqslant 700$

The corner points are \[F{\text{ }}\left( {0,{\text{ }}700} \right),{\text{ }}G{\text{ }}\left( {200,{\text{ }}600} \right),{\text{ }}H{\text{ }}\left( {400,{\text{ }}200} \right),{\text{ }}E{\text{ }}\left( {400,{\text{ }}0} \right).\]The values of Z at these corner points are as follows.

The maximum value of Z is \[1300\]which is attained at \[G{\text{ }}\left( {200,600} \right).\]Thus, the maximum profit obtained is Rs \[1300\]when \[200\]belts of type A and \[600\]belts of type B are produced.

13. Two Godowns X and Y have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shop A, B and C whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintals from the godowns to the shops are given in following table:

How should the supplies be transported to ration shops from godowns to minimize the transportation cost?

Let go down A supply x and y quintals of grain to the shops D and E respectively.

Then\[,{\text{ }}\left( {100{\text{ }} - {\text{ }}x{\text{ }} - {\text{ }}y} \right)\]will be supplied to shop F. 

The requirement at shop D is 60 quintals since x quintals are transported from godown A. Therefore, the remaining \[\left( {60{\text{ }} - x} \right)\]quintals will be transported from godown B.

Similarly, \[\left( {50{\text{ }} - {\text{ }}y} \right)\]quintals and \[40{\text{ }} - {\text{ }}\left( {100{\text{ }} - {\text{ }}x{\text{ }} - {\text{ }}y} \right){\text{ }} = {\text{ }}\left( {x{\text{ }} + {\text{ }}y{\text{ }} - {\text{ }}60} \right)\]quintals will be transported from go down B to shop E and F respectively.

The given problem can be represented diagrammatically as follows.

$x \geqslant 0,y \geqslant 0$, and $100 - x - y \geqslant 0$

$ \Rightarrow x \geqslant 0,y \geqslant 0$, and $x + y \leqslant 100$

$60 - x \geqslant 0,50 - y \geqslant 0$, and $x + y - 60 \geqslant 0$

$ \Rightarrow x \leqslant 60,y \leqslant 50$, and $x + y \geqslant 60$

Total transportation cost $z$ is given by,

$z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60)$

$ = 6x + 3y + 250 - 2.5x - 2.5y + 240 - 4x + 100 - 2y + 3x + 3y - 180$

$ = 2.5x + 1.5y + 410$

The given problem can be formulated as Minimize $z = 2.5x + 1.5y + 410 \ldots $ (1)

 subject to the constraints,

$x + y \leqslant 100$ ………..(II)

$x \leqslant 60$ ……………..(III)

$y \leqslant 50$ ……………..(iv)

$x + y \geqslant 60$ ………….(v)

$x,y \geqslant 0$ ……………..(vi)

The corner points are\[A{\text{ }}\left( {60,{\text{ }}0} \right),{\text{ }}B{\text{ }}\left( {60,{\text{ }}40} \right),{\text{ }}C{\text{ }}\left( {50,{\text{ }}50} \right),{\text{ }}and{\text{ }}D{\text{ }}(10,{\text{ }}50\]). The values of z at these corner points are as follows.

The minimum value of z is 510 at (10, 50).

Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.

14. An Aero plane can carry a maximum of 200 passengers. A profit of Rs. 400 is made on each first-class ticket and a profit of Rs. 300 is made on each second-class ticket. The airline reserves at least 20 seats for first class. However at least four times as many passengers prefer to travel by second class than by first class. Determine, how many tickets of each type must be sold to maximize profit for the airline.

Let there be x tickets of executive class and y tickets of economy class. Let Z be net profit of the airline.

 Here, we have to maximize z.

\[\begin{array}{*{20}{l}} {Z{\text{ }} = {\text{ }}500x\;{\text{ }}x{\text{ }}80/100{\text{ }} + {\text{ }}y{\text{ }}x{\text{ }}75/100} \\ {\;Z{\text{ }} = {\text{ }}400x{\text{ }} + {\text{ }}300y{\text{ }}....\left( I \right)\;} \end{array}\]

According to question 

\[x\; \geqslant \;20{\text{ }}....\left( {ii} \right)\;\]

 Also 

\[\begin{array}{*{20}{l}} {x{\text{ }} + {\text{ }}y\; \leqslant \;200{\text{ }}....\left( {iii} \right)\;} \\ {\; \Rightarrow \;x{\text{ }} + {\text{ }}4x\; \leqslant \;200\;} \\ { \Rightarrow \;5x\; \leqslant \;200\;} \\ {\; \Rightarrow \;x\; \leqslant \;40{\text{ }}....\left( {iv} \right)\;} \end{array}\]

Shaded region is feasible region having corner points \[A{\text{ }}\left( {20,{\text{ }}0} \right),{\text{ }}B{\text{ }}\left( {40,0} \right){\text{ }}C{\text{ }}\left( {40,{\text{ }}160} \right),{\text{ }}D{\text{ }}\left( {20,180} \right)\]

Now value of Z is calculated at corner point as

Hence, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the net profit of the airlines.

Yes, more passengers would prefer to travel by such an airline, because some amount of profit is invested for welfare fund.

15. A diet for a sick person must contain at least \[{\mathbf{4000}}\]units of vitamins, 50 units of minerals and \[{\mathbf{1400}}\]units of calories. Two foods A and B are available at a cost of Rs. 5 and Rs. 4 per unit respectively. One unit of food A contains 200 unit of vitamins, 1 unit of minerals and 40 units of calories whereas one unit of food B contains 100 units of vitamins, 2 units of minerals and 40 units of calories. Find what combination of the food A and B should be used to have least cost but it must satisfy the requirements of the sick person.

Let x units of food A

y units of food B

Where \[x \geqslant 0,{\text{ }}y \geqslant 0\]

We are to minimize

\[Z{\text{ }} = {\text{ }}5x{\text{ }} + {\text{ }}4y\]

\[200x{\text{ }} + {\text{ }}100y{\text{ }} \geqslant {\text{ }}4000\]

\[\begin{array}{*{20}{l}} {2x{\text{ }} + {\text{ }}y{\text{ }} \geqslant {\text{ }}40\;\;\;\;\;.....\left( 1 \right)} \\ {X{\text{ }} + {\text{ }}2y{\text{ }} \geqslant {\text{ }}50\;\;\;\;\;.....\left( 2 \right)} \\ {40x{\text{ }} + {\text{ }}40y{\text{ }} \geqslant {\text{ }}1400} \\ {X{\text{ }} + {\text{ }}y{\text{ }} \geqslant {\text{ }}35\;\;\;\;\;\;.......\left( 3 \right)} \end{array}\] Where, 

\[x{\text{ }} \geqslant {\text{ }}0,{\text{ }}y{\text{ }} \geqslant {\text{ }}0\]

From equation (1) to and we get,

\[\begin{array}{*{20}{l}} {2x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}40} \\ {If\;x = 0\;then\;y{\text{ }} = {\text{ }}40OY{\text{ }}in\;L{\text{ }}\left( {0,40} \right)} \\ {If\;y = 0\;then\;x{\text{ }} = 20OX{\text{ }}in\;A{\text{ }}\left( {20,0} \right)} \end{array}\]

From equation (2) to and we get,

\[\begin{array}{*{20}{l}} {X{\text{ }} + {\text{ }}2y{\text{ }} = {\text{ }}50} \\ {If\;x = 0\;then\;y = 25OY{\text{ }}in\;M{\text{ }}\left( {0,25} \right)} \\ {If\;y = 0\;then\;x = 50OX{\text{ }}in\;B{\text{ }}\left( {50,0} \right)} \end{array}\]

From equation (3) to and we get,

\[\begin{array}{*{20}{l}} {X{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}35} \\ {If\;x = 0\;then\;y = {\text{ }}35OY{\text{ }}in\;N{\text{ }}\left( {0,35} \right)} \\ {If\;y = 0\;then\;x = {\text{ }}35OX{\text{ }}in\;C{\text{ }}\left( {35,0} \right)} \end{array}\]

Shaded region is the feasible region, which is unbounded.

Corner point is

\[\begin{array}{*{20}{l}} {B{\text{ }}\left( {50,0} \right)} \\ {D{\text{ }}\left( {20,15} \right)} \\ {E{\text{ }}\left( {5,30} \right)} \\ {L{\text{ }}\left( {0,40} \right)} \end{array}\]

\[\begin{array}{*{20}{l}} {At{\text{ }}point\;B{\text{ }}\left( {50,0} \right) \Rightarrow Z = 5\left( {50} \right){\text{ }} + {\text{ }}4\left( 0 \right){\text{ }} = {\text{ }}250} \\ {At{\text{ }}point\;D{\text{ }}\left( {20,15} \right),{\text{ }}Z = 5\left( {20} \right){\text{ }} + {\text{ }}4\left( {15} \right){\text{ }} = {\text{ }}100 + 60 = 160} \\ {At{\text{ }}point\;E{\text{ }}\left( {5,30} \right),{\text{ }}Z = 5\left( 5 \right){\text{ }} + {\text{ }}4\left( {30} \right){\text{ }} = {\text{ }}25 + 120 = 145} \\ {At{\text{ }}point\;L{\text{ }}\left( {0,40} \right),{\text{ }}Z = 5\left( 0 \right){\text{ }} + {\text{ }}4\left( {40} \right){\text{ }} = {\text{ }}0 + 160 = 160} \\ {Then{\text{ }}at{\text{ }}least{\text{ }}cost\; = Rs145{\text{ }}at{\text{ }}\left( {5,30} \right)} \end{array}\]

Food A: 5 units, Food B: 30 units.

Significance of CBSE Class 12 Maths Chapter 12 Linear programming Important Questions

As mentioned earlier, this chapter holds immense importance in the Class 12 Maths syllabus as it introduces the concepts and principles of linear programming to the students. Preparing this chapter will need the complete study material along with the important questions suggested by the experts.

These questions have been formulated by following the latest CBSE syllabus to cover the use of all concepts and formulas taught in the chapter. Hence, adding these questions and answers to your practice sessions will be very fruitful in terms of conceptual development.

Benefits of CBSE Class 12 Maths Chapter 12 Linear programming Important Questions

Explore the advantages of delving into CBSE Class 12 Maths Chapter 12 - Linear Programming through its pivotal set of important questions. Discover how these questions can aid in grasping the core concepts, reinforcing knowledge, and ultimately mastering the complex principles of Linear Programming for enhanced understanding. Here are few points: 

Focus on key topics for efficient studying.

Prepares students for exams and reduces anxiety.

Reinforces understanding of fundamental concepts.

Teaches effective time management.

Enables self-assessment and progress tracking.

Strategic approach for higher scores.

Covers a wide range of topics for comprehensive understanding.

Supports exam preparation and boosts confidence.

Download CBSE Class 12 Maths Chapter 12 Linear programming Important Questions PDF

Get the PDF version of these questions and answers for free and learn how to solve such problems of linear programming faster. Use your study time efficiently by imbibing the answering styles given in the solutions to score more in the exams.

Reviewing all the crucial questions for Class 12 Maths Chapter 12 - Linear Programming provides students with a solid grasp of the chapter's topics. The extra and important questions for Class 12 Maths Chapter 12 - Linear Programming engage in a concept-focused discussion, encompassing all chapter themes. This question-and-answer method proves time-saving during exam prep, offering an efficient way to revise the chapter and enhance understanding. Practicing these important questions streamlines preparation and boosts confidence for the upcoming exams.

Important Related Links for CBSE Class 12 Maths 

FAQs on Important Questions for CBSE Class 12 Maths Chapter 12 - Linear Programming 2024-25

1. How can I learn the methods of solving linear programming questions?

Concentrate on the concepts and follow the solutions given for the NCERT exercises and important questions to learn these methods.

2. Is it important to practise the important questions of linear programming?

Solving these questions will broaden your preparation level for this chapter. You will learn new answering skills from the solutions.

3. Are these questions important for the Class 12 Maths exam?

These questions are suggested by the experts after following the CBSE syllabus and are important to practise.

4. How can I complete preparing Class 12 Maths Linear Programming?

Follow the theoretical explanation first and then solve the exercises. Proceed to solve the important questions and follow the solutions. This is how you can complete preparing this chapter.

5. Where can I get the solutions to these important questions?

The solutions are provided in the same file along with the important questions for your convenience. 

CBSE Class 12 Maths Important Questions

Cbse study materials.

NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming

Ncert solutions for class 12 maths chapter 12 – linear programming pdf.

NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 12 – Linear Programming Maths NCERT Solutions for Class 12 to help you to score more marks in your board exams and as well as competitive exams.

Share with friends: