NCERT Class 12 Mathematics Chapter 12 Linear Programming

NCERT Class 12 Mathematics Chapter 12 Linear Programming Solutions, NCERT Solutions For Class 12 Maths, NCERT Class 12 Mathematics Chapter 12 Linear Programming Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 12 Mathematics Chapter 12 Linear Programming Question Answer and select needs one.

NCERT Class 12 Mathematics Chapter 12 Linear Programming

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 12 Mathematics Chapter 12 Linear Programming Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Mathematics Chapter 12 Linear Programming Solutions for All Subject, You can practice these here.

Linear Programming

Chapter – 12

Solve the following Linear Programming Problems graphically:

1. Maximise Z = 3x + 4y

subject to the constraints : x + y ≤ 4 , x ≥ 0, y ≥ 0

Ans: The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0 and y ≥ 0 is given by

Since, the corner points of the feasible region are O(0, 0), A(4, 0) and B(0, 4) . The value of Z at these points are as follows:

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

2. Minimise Z = – 3x + 4y 

subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Ans: The feasible region determined by the system of constraints, x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0 and y ≥ 0, is given by

Since, the corner points of the feasible region are O(0, 0), A(4, 0), B(2, 3) and C(0,4). The value of Z at these corner points are as follows:

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

3. Maximise Z = 5x + 3y

subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

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

Since, the corner points of the feasible region are O(0, 0) , A(2, 0) B(0, 3) and

The value of Z at these corner points are as follows:

Corner point	Z = 5x + 3y
O(0,0)	0
A(2,0)	10
B(0,3)	9
		→Maximum

4. Minimise Z = 3x + 5y 

such that x + 3y ≥ 3 , x + y ≥ 2, x, y ≥ 0.

Ans: The feasible region determined by the system of constraints, x+3y≥3, x+y≥2 and x, y≥0, is given by

Since, the feasible region is unbounded, the corner points of the feasible region are

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

Corner point	Z = 3x + 5y
A(3,0)	9
	7	→Minimum
C(0,2)	10

As the feasible region is unbounded, therefore, 7 may or may not be the minimum value of Z. For this, we draw the graph of the inequality, 3x + 5y < 7, and check whether the resulting half plane has points in common with the feasible region or not.

Since, feasible region has no common point with 3x + 5y < 7

5. Maximise Z = 3x + 2y 

subject to x + 2y ≤ 10, 3x + y ≤ 15 , x, y ≥ 0.

Ans: The feasible region determined by the constraints, x + 2y ≤ 10, 3x + y ≤ 15 and x, y ≥ 0 , is given by

Since the corner points of the feasible region are A(5, 0), B(4, 3) and C(0,5). The values of Z at these corner points are as follows:

Thus, the maximum value of Z is 18 at the point B(4, 3).

6. Minimise Z = x + 2y

subject to 2x + y ≥ 3 , x + 2y ≥ 6, x, y ≥ 0.

Ans: The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6 and x, y ≥ 0, is given by

Since the corner points of the feasible region are A(6, 0) and B(0, 3). The values of Z at these corner points are as follows:

Since the values of Z at points A and B is same. If we take any other point such as (2,2) on line x + 2y = 6 , then Z = 6

Thus, the minimum value of Z occurs for more than 2 points.

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

Show that the minimum of Z occurs at more than two points.

7. Minimise and Maximise Z = 5x + 10y 

subject to x + 2y ≤ 120, x + y ≥ 60 , x – 2y ≥ 0 , x, y ≥ 0

Ans: The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0 and x, y ≥ 0 is given by

Since the corner points of the feasible region are A(60, 0), B(120, 0), C(60,30) and D(40, 20) The values of Z at these corner points are as follows:

The minimum value of Z is 300 at A(60, 0) and the maximum value of Z is 600 at all the points on the line segment joining B(120, 0) and C(60,30).

8. Minimise and Maximise Z = x + 2y 

subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0

Ans: The feasible region determined by the constraints, x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200 and x, y ≥ 0 is given by

The corner points of the feasible region are A(0, 50) B(20, 40) C(50,100) and D(0, 200) The values of Z at these corner points are as follows:

The maximum value of Z is 400 at D(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).

9. Maximise Z = – x + 2y, 

subject to the constraints: x ≥ 3 , x + y ≥ 5 , x + 2y ≥ 6, y ≥ 0

Ans: The feasible region determined by the constraints, x ≥ 3, x + y ≥ 5, x + 2y ≥ 6 and y ≥ 0 is given by

Since, the feasible region is unbounded, the values of Z at corner points A(6, 0), B(4, 1) and C(3,2) are as follows:

As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value. For this, we graph the inequality, – x + 2y > 1 , and check whether the resulting half plane has points in common with the feasible region or not.

The resulting feasible region has points in common with the feasible region.

Thus, Z = 1 is not the maximum value.

Therefore, Z has no maximum value.

10. Maximise Z = x + y, 

subject to x – y ≤ – 1 , – x + y ≤ 0 , x, y ≥ 0

Ans: The feasible region determined by the constraints, x – y ≤ -1,- x + y ≤ 0 and x, y ≥ 0, is given by

There is no feasible region and thus, Z has no maximum value.