NCERT Solution Class 12 Chapter 12 - Linear Programming Exercise 12.1

NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1 – Free PDF Download

Exercise 12.1 of NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming is based on the following topics:

1. Introduction
2. Linear Programming Problem and its Mathematical Formulation
   1. Mathematical formulation of the problem
   2. Graphical method of solving linear programming problems

Solve all the problems of this exercise to get thorough with the concepts and topics covered in the entire chapter.

NCERT Solutions for Class 12 Maths Chapter 12- Linear Programming Exercise 12.1

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Access Other Exercises of Class 12 Maths Chapter 12

Exercise 12.2 Solutions 11 Questions

Miscellaneous Exercise On Chapter 12 Solutions 10 Questions

Access Answers to NCERT Class 12 Maths Chapter 12 Exercise 12.1

1. Maximise Z = 3x + 4y

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

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

2. Minimise Z = −3x + 4y

subject to.

Solution:

The feasible region determined by the system of constraints,
is 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

Hence, the minimum value of Z is – 12 at the point (4, 0)

3. Maximise Z = 5x + 3y

subject to.

Solution:

The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are 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

Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19)

4. Minimise Z = 3x + 5y

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

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 the resulting half plane have 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 = 3x + 2y

subject to.

Solution:

The feasible region determined by the constraints, x + 2y ≤ 10, 3x + 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

Hence, the maximum value of Z is 18 at the point (4, 3)

6. Minimise Z = x + 2y

subject to
.

Solution:

The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 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

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 = 5x + 10y

subject to.

Solution:

The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 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

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 + 2y

subject to.

Solution:

The feasible region determined by the constraints, x + 2y ≥ 100, 2x − y ≤ 0, 2x + 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

The maximum value of Z is 400 at point (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 + 2y, 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

Since the feasible region is unbounded, hence, 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, is given below

There is no feasible region and therefore, z has no maximum value.

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NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12

NCERT Solutions