NCERT Solutions 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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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

Class 12 Maths Chapter 12 : Exercises Breakdown

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.