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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

NCERT Solutions for Class 12 Maths Chapter 12 are a detailed explanation of the problems the chapter includes. Physics Wallah experts have come up with simply laid out solutions that shall help students of Class 12 CBSE understand mathematical concepts. The article here is intended to be a ready reference for students that shall help them understand Linear Programming.
authorImageJasdeep Bhatia25 Jan, 2024
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NCERT Solutions for Class 12 Maths Chapter 12

NCERT Solutions for Class 12 Maths Chapter 12

NCERT Solutions for Class 12 Maths Chapter 12 : The solutions given here help form a clear understanding of Linear Programming Class 12 . As a helpful reference tool, the article makes it easier for students to grasp the concepts of Linear Programming. The best method that can be used for achieving the best outcome in a mathematical model where linear relationships explain represent the requirements is called Linear Programming. This article helps students learn the fundamentals of Linear Programming much better. Repeated practice of the solutions given here by Physics Wallah experts help students ace the board exams. This article’s Maths Class 12 Chapter 12 Linear Programming solutions are given in easy to understand steps that help form a clear concept of Linear Programming. The Physics Wallah experts have taken care to see to it that the simplest methods are used in explaining the problems. As such, this article shall be of great help to even the weaker students of Linear Programming. Linear Programming Class 12 becomes a much easier challenge to face with our Class 12 Maths Chapter 12 Linear Programming solutions . The solutions are in accordance with the Class 12 CBSE syllabus. Physics Wallah also gives students the option to download the PDF file free so they can refer to it offline whenever they study.

NCERT Solutions for Class 12 Maths Chapter 12 Overview

The experts at Physics Wallah seek to help students clarify their doubts about Linear Programming Class 12 with the solutions that this article contains. The experts have put themselves in the weaker students’ shoes and that has helped keep the solutions very simple. The methods used to solve the problems are indeed very easy to understand. The Class 12 Maths Chapter 12 Linear Programming solutions Class 12 Maths Chapter 12 Linear Programming solutions given here help students overcome their doubts about Linear Programming. Students can always refer to this article, available both online and offline and practice regularly so that they can improve their skills at solving Linear Programming problems. The Linear Programming Class 12 solutions from Physics Wallah help students a lot when it comes to forming a clear and thorough understanding of the subject. Students have for long found this chapter confusing. The easy solutions given in this article help even the weaker students understand Linear Programming deeply since every aspect is discussed in details.

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

Exercise 12.1 Page No. 513 1. Maximise Z = 3 x + 4 y Subject to the constraints: Solution: The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is given below. O (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of Z at these points are given below.
Corner point Z = 3x + 4y
O (0, 0) 0
A (4, 0) 12
B (0, 4) 16 Maximum
Hence, the maximum value of Z is 16 at the point B (0, 4). 2. Minimise Z = −3 x + 4 y subject to The feasible region determined by the system of constraints, given below. O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region. The values of Z at these corner points are given below.
Corner point Z = – 3x + 4y
O (0, 0) 0
A (4, 0) -12 Minimum
B (2, 3) 6
C (0, 4) 16
Hence, the minimum value of Z is – 12 at the point (4, 0). 3. Maximise Z = 5 x + 3 y subject to Solution: The feasible region determined by the system of constraints, 3 x + 5 y ≤ 15, 5 x + 2 y ≤ 10, x ≥ 0, and y ≥ 0, is given below. O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are the corner points of the feasible region. The values of Z at these corner points are given below.
Corner point Z = 5x + 3y
O (0, 0) 0
A (2, 0) 10
B (0, 3) 9
C (20 / 19, 45 / 19) 235 / 19 Maximum
Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19). 4. Minimise Z = 3 x + 5 y such that Solution: The feasible region determined by the system of constraints, x + 3y ≥ 3, x + y ≥ 2, and x, y ≥ 0, is given below. It can be seen that the feasible region is unbounded. The corner points of the feasible region are A (3, 0), B (3 / 2, 1 / 2) and C (0, 2). The values of Z at these corner points are given below.
Corner point Z = 3x + 5y
A (3, 0) 9
B (3 / 2, 1 / 2) 7 Smallest
C (0, 2) 10
7 may or may not be the minimum value of Z because the feasible region is unbounded. For this purpose, we draw the graph of the inequality, 3x + 5y < 7 and check whether the resulting half-plane has common points with the feasible region or not. Hence, it can be seen that the feasible region has no common point with 3x + 5y < 7. Thus, the minimum value of Z is 7 at point B (3 / 2, 1 / 2). 5. Maximise Z = 3 x + 2 y subject to Solution: The feasible region determined by the constraints, x + 2 y ≤ 10, 3 x + y ≤ 15, x ≥ 0, and y ≥ 0, is given below. A (5, 0), B (4, 3), C (0, 5) and D (0, 0) are the corner points of the feasible region. The values of Z at these corner points are given below.
Corner point Z = 3x + 2y
A (5, 0) 15
B (4, 3) 18 Maximum
C (0, 5) 10
Hence, the maximum value of Z is 18 at points (4, 3). 6. Minimise Z = x + 2 y subject to Solution: The feasible region determined by the constraints, 2 x + y ≥ 3, x + 2 y ≥ 6, x ≥ 0, and y ≥ 0, is given below. A (6, 0) and B (0, 3) are the corner points of the feasible region. The values of Z at the corner points are given below.
Corner point Z = x + 2y
A (6, 0) 6
B (0, 3) 6
Here, the values of Z at points A and B are same. If we take any other point, such as (2, 2) on line x + 2y = 6, then Z = 6. Hence, the minimum value of Z occurs for more than 2 points. Therefore, the value of Z is minimum at every point on the line x + 2y = 6. 7. Minimise and Maximise Z = 5 x + 10 y subject to Solution: The feasible region determined by the constraints, x + 2 y ≤ 120, x + y ≥ 60, x − 2 y ≥ 0, x ≥ 0, and y ≥ 0, is given below. A (60, 0), B (120, 0), C (60, 30), and D (40, 20) are the corner points of the feasible region. The values of Z at these corner points are given below.
Corner point Z = 5x + 10y
A (60, 0) 300 Minimum
B (120, 0) 600 Maximum
C (60, 30) 600 Maximum
D (40, 20) 400
The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on the line segment joining (120, 0) and (60, 30). 8. Minimise and Maximise Z = x + 2 y subject to Solution: The feasible region determined by the constraints, x + 2 y ≥ 100, 2 x y ≤ 0, 2 x + y ≤ 200, x ≥ 0, and y ≥ 0, is given below. A (0, 50), B (20, 40), C (50, 100) and D (0, 200) are the corner points of the feasible region. The values of Z at these corner points are given below.
Corner point Z = x + 2y
A (0, 50) 100 Minimum
B (20, 40) 100 Minimum
C (50, 100) 250
D (0, 200) 400 Maximum
The maximum value of Z is 400 at points (0, 200), and the minimum value of Z is 100 at all the points on the line segment joining the points (0, 50) and (20, 40). 9. Maximise Z = − x + 2 y , subject to the constraints. Solution: The feasible region determined by the constraints, is given below. Here, it can be seen that the feasible region is unbounded. The values of Z at corner points A (6, 0), B (4, 1) and C (3, 2) are given below.
Corner point Z = – x + 2y
A (6, 0) Z = – 6
B (4, 1) Z = – 2
C (3, 2) Z = 1
Since the feasible region is unbounded, z = 1 may or may not be the maximum value. For this purpose, we graph the inequality, – x + 2y > 1, and check whether the resulting half-plane has points in common with the feasible region or not. Here, the resulting feasible region has points in common with the feasible region. Hence, z = 1 is not the maximum value. Z has no maximum value. 10. Maximise Z = x + y , subject to Solution: The region determined by the constraints There is no feasible region, and therefore, z has no maximum value.

Exercise 12.2 page no: 519

1. Reshma wishes to mix two types of food, P and Q, in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A, and 11 units of vitamin B. Food P costs Rs 60/kg, and Food Q costs Rs 80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B. Determine the minimum cost of the mixture? Solution: Let the mixture contain x kg of food P and y kg of food Q. Hence, x ≥ 0 and y ≥ 0 The given information can be compiled in a table, as given below.
Vitamin A (units/kg) Vitamin B (units/kg Cost (Rs/kg)
Food P 3 5 60
Food Q 4 2 80
Requirement (units/kg) 8 11
The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Hence, the constraints are 3 x + 4 y ≥ 8 5 x + 2 y ≥ 11 The total cost of purchasing food is Z = 60x + 80y. So, the mathematical formulation of the given problem can be written as Minimise Z = 60x + 80y (i) Now, subject to the constraints, 3 x + 4 y ≥ 8 … (2) 5 x + 2 y ≥ 11 … (3) x , y ≥ 0 … (4) The feasible region determined by the system of constraints is given below. Clearly, we can see that the feasible region is unbounded. A (8 / 3, 0), B (2, 1 / 2) and C (0, 11 / 2) The values of Z at these corner points are given below.
Corner point Z = 60x + 80 y
A (8 / 3, 0) 160 Minimum
B (2, 1 / 2) 160 Minimum
C (0, 11 / 2) 440
Here, the feasible region is unbounded; therefore, 160 may or may not be the minimum value of Z. For this purpose, we graph the inequality, 60 x + 80 y < 160 or 3 x + 4 y < 8, and check whether the resulting half-plane has points in common with the feasible region or not. Here, it can be seen that the feasible region has no common point with 3x + 4y < 8. Hence, at the line segment joining the points (8 / 3, 0) and (2, 1 / 2), the minimum cost of the mixture will be Rs 160. 2. One kind of cake requires 200g flour and 25g of fat, and another kind of cake requires 100g of flour and 50g 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? Solution: Let the first kind of cake be x and second kind of cakes be y. Hence, x ≥ 0 and y ≥ 0 The given information can be compiled in a table, as shown below.
Flour (g) Fat (g)
Cakes of the first kind, x 200 25
Cakes of the second kind, y 100 50
Availability 5000 1000
So, 200x + 100y ≤ 5000 2x + y ≤ 50 25x + 50y ≤ 1000 x + 2y ≤ 40 The total number of cakes Z that can be made are Z = x + y The mathematical formulation of the given problem can be written as Maximise Z = x + y (i) Here, subject to the constraints, 2x + y ≤ 50 (ii) x + 2y ≤ 40 (iii) x, y ≥ 0 (iv) The feasible region determined by the system of constraints is given below. A (25, 0), B (20, 10), O (0, 0) and C (0, 20) are the corner points. The values of Z at these corner points are as given below.
Corner point Z = x + y
A (25, 0) 25
B (20, 10) 30 Maximum
C (0, 20) 20
O (0,0) 0
Hence, the maximum numbers of cakes that can be made are 30 (20 cakes of one kind and 10 cakes of other kind).

Topics Included in NCERT Solutions for Class 12 Maths Chapter 12

The following Linear Programming topics have been included in this article in accordance with the Linear Programming guidelines mentioned by NCERT.

12.1 Introduction

12.2 Linear Programming Problem and its Mathematical Formulation

12.2.1 Mathematical formulation of the problem

12.2.2 Graphical method of solving linear programming problems

12.3 Different Types of Linear Programming Problems

12.3.1 Manufacturing problems

12.3.2 Diet problems

12.3.3 Transportation problems

NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming

The chapter on Linear Programming is a complete unit carrying five marks of the total marks of eighty. With the miscellaneous exercise, there are two exercises as well. These shall help students form a clear understanding of the concept of Linear Programming. This article on Linear Programming Class 12 discusses the following—
  • 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 or unbounded),
  • Feasible and infeasible solutions, and
  • Optimal feasible solutions (up to three non-trivial constraints)

NCERT Solutions for Class 12 Maths Chapter 12- Key Features of Linear Programming

This article on Class 12 Maths Chapter 12 Linear Programming solutions helps students gain a clear understanding of the vital concepts of the subject. It touches upon the key concepts including,
  • Diet problems
  • Manufacturing problems
  • Transportation problems
The article explains the important concepts of feasible region and feasible solutions. The method that has been used to solve the Linear Programming problems is the Corner Point Method. The question of optimal solution has also been discussed.

NCERT Solutions for Class 12 Maths Chapter 12 FAQs

What is the concept of Linear Programming? 

Linear Programming involves subjecting mathematical functions to constraints which are then maximized or minimized accordingly. This chapter covers objective functions and constraints that are completely linear. The linear functions’ maximum and minimum values are plotted on the xy coordinate for further deductions. Optimising the mathematical functions exposed to linear constraints happens to be the main objective of Linear Programming. 

How can I access the NCERT Solutions for Class 12 Maths Chapter 11?

You’ll simply have to visit the Physics Wallah website from where you can download the entire chapter. All the solutions given here have been worked on by experts at Physics Wallah and they are very helpful in understanding Linear Programming. 

How does this article help students? 

Students can refer to Linear Programming Class 12 solutions included in this article and clear their doubts about the subject. Coupled with regular practice, referring to this article goes a long way in acing Linear Programming. 

How much do Class 12 Maths Chapter 12 Linear Programming solutions help students score well in the board exams? 

The solutions given here are simple and easy to understand. Well explained solutions to the problems like the ones included here do help score higher marks in the board exams. Of course, regular practice is a must. 

How can I form a deep and thorough understanding of Linear Programming? 

While referring to this article and regular practice of the solutions help, to go deep into the subject, students shall have to study the textbook thoroughly. Only doing so can help form a strong foundation on Linear Programming. 
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