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NCERT Solutions Class 12 Maths Chapter 12 Linear Programming

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NCERT Maths Chapter 12 Linear Programming Class 12 Solutions - FREE PDF Download

NCERT Solutions for Class 12 Maths Chapter 12: Linear Programming explains the various concepts and principles of linear programming. Students will learn these concepts and solve the problems given in the exercise to build their concepts well. To make this process easier, download and refer to the CBSE Class 12 Maths Chapter 12 Linear Programming NCERT Solutions developed by the experts.

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Table of Content
1. NCERT Maths Chapter 12 Linear Programming Class 12 Solutions - FREE PDF Download
2. Glance on Maths Chapter 12 Class 12 - Linear Programming 
3. Access Exercise wise NCERT Solutions for Class 12 Maths Chapter 12
4. What is Linear Programming?
5. Exercises Under NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming
6. Access NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming
7. Overview of Deleted Syllabus for CBSE Class 12 Maths Linear Programming
8. Class 12 Maths Chapter 12 : Exercises Breakdown
9. Other Study Material for CBSE Class 12 Maths Chapter 12
10. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs


In this chapter, you will learn how to plot linear equations on a graph. Since many applications in Mathematics involve systems of inequalities or equations, this chapter will teach you how to apply these inequalities or equations.


Glance on Maths Chapter 12 Class 12 - Linear Programming 

  • Linear Programming chapter dives into optimization problems and how to solve them using linear programming techniques.

  • Graphical Method of Solving Linear Programming Problems involves plotting the constraints on a graph to find the feasible region, which is the area satisfying all constraints. The corner points of this region are then evaluated using the objective function to find the optimal solution (maximum or minimum value).

  • Explore Different Types of Linear Programming Problems variations of linear programming problems, such as problems with mixed constraints (both equations and inequalities) and minimization problems.

  • This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 12 - Linear Programming, which you can download as PDFs.

  • There is one exercise (10 fully solved questions) in class 12th maths chapter 12 Linear Programming.


Access Exercise wise NCERT Solutions for Class 12 Maths Chapter 12

Current Syllabus Exercises of Class 12 Maths Chapter 12

NCERT Solutions of Class 10 Maths Linear Programming Exercise 12.1


What is Linear Programming?

Linear programming is a method to optimise any problem with several variables and constraints. It is a simple but powerful tool every data scientist should be aware of. The linear programming method is used to solve optimisation problems where all the constraints, along with the objective function, are linear equalities or inequalities

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Exercises Under NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

Chapter 12 of NCERT Class 12 Maths deals with Linear Programming. Linear Programming is a technique used to find the optimal solution to a problem where we need to maximize or minimize a linear function subject to a set of constraints that are also linear in nature.


Exercise 12.1: This exercise consists of 10 questions that are based on formulating linear programming problems and solving them graphically. These questions help students understand the basic concepts and steps involved in solving linear programming problems.


Access NCERT Solutions for Class 12 Maths Chapter 12 – Linear Programming

Exercise 12.1

1. Maximize \[\mathbf{Z=3x+4y}\]

Subject to the constraints: \[\mathbf{x+y\le 4}\], \[\mathbf{x\ge 0}\], \[\mathbf{y\ge 0}\]

Ans:  The given constraints are, \[x+y\le 4\], \[x\ge 0\], \[y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points O (0, 0), A (4, 0), and B (0, 4) at the corners


Feasible Region having points O (0, 0), A (4, 0), and B (0, 4) at the corners


The points at the corners in the feasible region are O (0, 0), A (4, 0), and B (0, 4). Z assumes the following values on these points.

Corner Point

Z = 3x + 4y


O(0,0)

0


A(4,0)

12


B(0,4)

16

→ Maximum

Table of values

Thus, the maximum value of Z is 16 at the point B (0,4).


2. Minimize Z=-3x+4y subject to \[\mathbf{x+2y\le 8}\], \[\mathbf{3x+2y\le 12}\], \[\mathbf{x\ge 0}\], \[\mathbf{y\ge 0}\]

Ans:  The given constraints are, \[x+2y\le 8\], \[3x+2y\le 12\], \[x\ge 0\]and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points O (0, 0), A (4, 0), B (2, 3), and C (0, 4) at the corners


Feasible Region having points O (0, 0), A (4, 0), B (2, 3), and C (0, 4) at the corners


The points at the corners in the feasible region are  O (0, 0), A (4, 0), B (2, 3), and C (0, 4). Z assumes the following values on these points.

Corner Point

Z=-3x+4y


O(0,0)

0


A(4,0)

-12

-> Minimum

B(2,3)

6


C(4,0)

16


Thus, the minimum value of Z is -12 at the point (4, 0).


3.Maximize \[\mathbf{Z=5x+3y}\] subject to \[\mathbf{3x+5y\le 15}\], \[\mathbf{5x+2y\le 10}\], \[\mathbf{x\ge 0}\], \[\mathbf{y\ge 0}\].

Ans: The given constraints are,\[3x+5y\le 15\], \[5x+2y\le 10\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points O (0, 0), A (2, 0), B (0, 3), at the corners


Feasible Region having points O (0, 0), A (2, 0), B (0, 3), at the corners


The points at the corners in the feasible region are O (0, 0), A (2, 0), B (0, 3), and \[C\left( \frac{20}{19},\frac{45}{19} \right)\]. Z assumes the following values on these points.

Corner Point

Z=5x+3y


O(0,0)

0


A(2,0)

10


B(0,3)

9


C(20/19,45/19)

235/18

-> Maximum

Thus, the maximum value of Z is \[\frac{235}{19}\] at the point \[\left( \frac{20}{19},\frac{45}{19} \right)\].


4.Minimize \[Z=3x+5y\]such that \[\mathbf{x+3y\ge 3}\], \[\mathbf{x+y\ge 2}\], \[\mathbf{x,y\ge 0}\]

Ans: The given constraints are, \[x+3y\ge 3\], \[x+y\ge 2\], and \[x,y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points A (0, 2), B (3, 0), at the corners


Feasible Region having points A (0, 2), B (3, 0), at the corners


We can see that the feasible region is not bounded.

The points at the corners in the feasible region are A (3, 0), \[B\left( \frac{3}{2},\frac{1}{2} \right)\] , and C (0, 2). Z assumes the following values on these points.

Corner Point

Z=3x+5y


A(3,0)

9


B(3/2,½)

7

->Smallest

C(0,2)

10


Since the feasible region is unbounded, we cannot be sure that 7 is the minimum value of Z. To confirm this, we need to sketch the graph of the inequality, \[3x+5y<7\], and see if the resulting plane has any point in common with the feasible region.

From the graph that we sketched, we can see that there is no common point between feasible regions and the sketched inequality \[3x+5y<7\].

 Z achieves minimum value 7  at \[\left( \frac{3}{2},\frac{1}{2} \right)\].


5.Maximize Z=3x+2y subject to \[\mathbf{x+2y\le 10}\], \[\mathbf{3x+y\le 15}\], \[\mathbf{x,y\ge 0}\].

Ans: The given constraints are,  \[x+2y\le 10\], \[3x+y\le 15\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points A (5, 0), B (4, 3), and C (0, 5) at the corners


Feasible Region having points A (5, 0), B (4, 3), and C (0, 5) at the corners


The points at the corners in the feasible region are A (5, 0), B (4, 3), and C (0, 5). Z assumes the following values on these points.

Corner Point

Z=3x+2y


A(5,0)

15


B(4,3)

18

->Maximum

C(0,5)

10


Z achieves maximum value 18 at (4,3).


6.Minimize \[Z=x+2y\]subject to \[2x+y\ge 3\], \[x+2y\ge 6\], \[x,y\ge 0\].

Ans:The given constraints are, \[2x+y\ge 3\], \[x+2y\ge 6\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points A (6, 0) and B (0, 3) at the corners


Feasible Region having points A (6, 0) and B (0, 3) at the corners


The points at the corners in the feasible region are A (6, 0) and B (0, 3). Z assumes the following values on these points.

Corner Point

Z=x+2y


A(6,0)

6


B(0,3)

6

->Maximum

We can see that the value of Z is the same on both A and B hence, we will need to check on the other point on the line \[x+2y=6\]as well. The value of Z is 6 at point (2,2) also. Hence, the minimum value of Z Occurs at more than two points

Thus, the value of Z is minimum at every point on the line, \[x+2y=6\]


7. Minimize and Maximize Z=5x+10y subject to\[\mathbf{x+2y\le 120}\], \[\mathbf{x+y\ge 60}\], \[\mathbf{x-2y\ge 0}\], \[\mathbf{x,y\ge 0}\]. 

Ans:The given constraints are,  \[x+2y\le 120\], \[x+y\ge 60\], \[x-2y\ge 0\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points A (60, 0), B (120, 0), C (60, 30), and D (40, 20) at the corners


Feasible Region having points A (60, 0), B (120, 0), C (60, 30), and D (40, 20) at the corners


The points at the corners in the feasible region are A (60, 0), B (120, 0), C (60, 30), and D (40, 20). Z assumes the following values on these points.

Corner Point

Z= 5x+10y


A(60, 0)

300

→Minimum

B(120, 0)

600

→Maximum

C(60, 30)

600

→Maximum

D(40,20)

400


Z achieves maximum and minimum values as 600 and 300 respectively. The point of maximum value is all the points on the line segment joining (120, 0) and (60, 30) and minimum value is (60,0).


8.Minimize and Maximize Z=x+2ysubject to \[\mathbf{x+2y\ge 100}\], \[\mathbf{2x-y\le 0}\], \[\mathbf{2x+y\le 200}\], \[\mathbf{x,y\ge 0}\]. 

Ans:The given constraints are, \[x+2y\ge 100\], \[2x-y\le 0\], \[2x+y\le 200\], \[x\ge 0\], and \[y\ge 0\], and the feasible region which is in accordance with the given constraints is 


Feasible Region having points A(0, 50), B(20, 40), C(50, 100), and D(0, 200) at the corners


Feasible Region having points A(0, 50), B(20, 40), C(50, 100), and D(0, 200) at the corners


The points at the corners in the feasible region are A(0, 50), B(20, 40), C(50, 100), and D(0, 200). Z assumes the following values on these points.

Corner Point

Z=x+2y


A(0, 50)

100

→Minimum

B(20,40)

100

→Minimum

C(50,100)

250


D(0,200)

400

→Maximum

Z achieves maximum and minimum values as 400 and 100 respectively. The point of maximum value is (0,200) and minimum value is all points on the line joining the points (0, 50) and (20, 40).


9. Maximize Z=-x+2y, subject to the constraints: \[\mathbf{x\ge 3}\], \[\mathbf{x+y\ge 5}\], \[\mathbf{x+2y\ge 6}\], \[\mathbf{y\ge 0}\]. 

Ans: The given constraints are, \[x\ge 3\], \[x+y\ge 5\], \[x+2y\ge 6\], and \[y\ge 0\]  and the feasible region which is in accordance with the given constraints is


Feasible Region having points A (6, 0), B (4, 1), and C (3, 2) at the corners


Feasible Region having points A (6, 0), B (4, 1), and C (3, 2) at the corners


We can see that the feasible region is not bounded. 

The points at the corners in the feasible region are A (6, 0), B (4, 1), and C (3, 2) . Z assumes the following values on these points.

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, we cannot be sure that 1 is the maximum value of Z. To confirm this, we need to sketch the graph of the inequality,  \[-x+2y>1\], and see if the resulting plane has any point in common with the feasible region.

From the graph that we sketched, we can see that there are common points between feasible regions and the sketched inequality. Thus, Z = 1 is not the maximum value. Z has no maximum value.


10. Maximize Z=x+y , subject to\[\mathbf{x-y\le -1}\], \[\mathbf{-x+y\le 0}\], \[\mathbf{x,y\ge 0}\]. 

Ans: The given constraints are, \[x-y\le -1\], \[-x+y\le 0\], \[x,y\ge 0\]and the feasible region which is in accordance with the given constraints is 


Graph having No feasible region


Graph having No feasible region


We can see from the graph that there is no feasible region, hence Z has no maximum value.


Overview of Deleted Syllabus for CBSE Class 12 Maths Linear Programming

Chapter

Dropped Topics

Linear Programming

12.3 Different Types of Linear Programming Problems 

Page 528-529 Summary Points 2-9


Class 12 Maths Chapter 12 : Exercises Breakdown

Exercise

Number of Questions

Exercise 12.1

10 Questions (All Questions Require Graphs and Long Questions)


Conclusion

Chapter 12  Linear Programming of Class 12 Maths, focuses on the formulation and graphical solution of linear programming problems (LPP). It emphasizes the practical applications of optimizing a linear objective function subject to a set of linear inequalities or constraints. This chapter is crucial for understanding how mathematical techniques can be applied to real-world problems involving optimization. From previous year's question papers, around 4 questions are typically asked from this chapter. Understanding and practising these solutions will help students score well on their exams.


Other Study Material for CBSE Class 12 Maths Chapter 12


Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


FAQs on NCERT Solutions Class 12 Maths Chapter 12 Linear Programming

1. How many sums are there in the NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming?

There are 3 exercises in the NCERT Class 12 Maths Chapter 12. In the first exercise, Ex.-12.1, there are 10 sums. In the second exercise, Ex.-12.2, there are 11 sums, and in the miscellaneous exercise, there are 10 sums. All the sums in this chapter are solved with stepwise explanations in the NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming.

2. What is meant by Linear Programming in mathematics?

In linear programming, mathematical functions are subjected to various constraints and thereby maximised or minimised accordingly. All the constraints and the objective functions dealt with in this chapter are linear. The maximum and minimum values of the linear functions are plotted on the xy coordinate for further deductions. The main purpose of linear programming is to optimise various mathematical functions in terms of linear constraints.

3. Can I get NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming online?

Yes, you can get the NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming on Vedantu. The solutions to all the sums are prepared by our subject matter experts for the reference of students. You can download and refer to the NCERT solutions to get a clear understanding of the best problem-solving techniques for the sums of linear programming. The stepwise solutions facilitate a self-study session as you can compare your answers with these expert solutions.

4. Can I download the NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming for free?

Yes, you can download the NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming for free from Vedantu. These NCERT Solutions are available in a PDF format and you can download the same from our mobile application as well. This feature of free downloadable NCERT Solutions of Vedantu enables all students with an internet connection to refer to the best study materials online.

5. What are the fundamental theorems for linear programming?

Theorem 1

Let us consider  R as the feasible region for linear programming and let's consider z to be ax+by, which is the objective function. When z reaches an optimal value, whether it be maximum or minimum and the variables x and y are the subjects to constraints. This optimal value that occurs must occur at a corner point of the said feasible region.


Theorem 2

Let us consider  R as the feasible region for the given linear programming problem and let z be ax+by, which is the objective function. When the R is bounded then z attains both the maximum and the minimum value of R where each recurs at the joint corner of JR. 

6. What are the steps followed to apply the corner point method?

The steps used in applying the corner point method are as follows:

  • First, it is important to find the feasible regions in the given linear programming problem, and along with it determine the corner point. This can be done by inspection or by solving the two equations of the line that intersect at that point.

  • The next step evolves the evaluation of the objective function, ie, z= ax+by at each of the corner points. Denote M and m respectively as the largest and smallest values of the given points.

  • When the feasible regions are bounded, then M and m respectively attain the maximum and the minimum of the objective function at the corner points. 

7. What is a feasible region?

A feasible region attributed to the common region is determined by all the constraints which include even the non-negative constraints x,y>0 of linear programming. All other regions that do not fall under the feasible regions are termed as the non-feasible regions. 

8. What is an objective function and a constraint?

An objective function refers to the linear function, for instance, z=px+py, where p and q are the constants. This linear equation that needs to be maximised or minimised is referred to as the objective function. When restrictions or linear inequalities are placed on a variable in the linear programming problem it is known as constraints. It is important for the student to familiarise and be thorough with these concepts in order to be able to do well in the examination. 

9. How to prepare for Class 12 Maths Chapter 12?

A subject such as Math demands a high amount of practice to be thorough with the concepts. With adequate and regular practice, the student will have all the important points at the tip of their fingers.  Therefore, without any doubt, the student needs to practise all the exercises that this chapter entails. To add the icing to the cake, the students can turn towards the NCERT Solutions for Class 12 Maths Chapter 12 available on the Vedantu website and on the Vedantu app at free of cost. These solutions will prove to be the best source of guidance for the students making them exam ready.

10. What are the key components of a Linear Programming Problem (LPP) explained in Linear Programming Class 12 NCERT Solutions?

The key components of an LPP include:

  • Objective Function: A linear function that needs to be maximized or minimized.

  • Constraints: A set of linear inequalities or equations that define the feasible region.

  • Feasible Region: The region in the graph that satisfies all constraints.

  • Decision Variables: The variables that impact the objective function and are subject to the constraints.

11. What is Linear Programming in Linear Programming Class 12 Solutions PDF?

Linear Programming (LP) in Linear Programming Class 12 Solutions PDF is a method to achieve the best outcome in a mathematical model with linear relationships. It involves optimizing a linear objective function, subject to linear equality and inequality constraints.

12 What is the graphical method of solving Linear Programming Problems in Chapter 12 Maths Class 12?

The graphical method involves:

  • Plotting each constraint on a graph.

  • Identifying the feasible region where all constraints overlap.

  • Finding the corner points (vertices) of the feasible region.

  • Evaluating the objective function at each corner point to find the optimal solution.

13. How do you formulate a Problem in linear programming class 12 solutions?

Formulating an LPP in Class 12 linear programming involves:

  • Defining the decision variables.

  • Writing down the objective function.

  • Stating the constraints.

  • Ensuring all functions and constraints are linear.