These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1/
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1
Ex 12.1 Class 12 Question 1.
Maximize Z = 3x + 4y
subject to the constraints:
x + y ≤ 4, x ≥ 0, y ≥ 0.
Solution:
The objective function is Z = 3x + 4y
The constraints are x + y ≤ 4, x ≥ 0, y ≥ 0.
∴ Maximum value of Z is 16 at B (0, 4).
Question 2.
Minimize Z = – 3x + 4y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Solution:
The objective function is Z = – 3x + 4y The constraints are x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
∴ Maximum value of Z is – 12 at A (4, 0).
Question 3.
Maximize Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Solution:
The objective function is Z = 5x + 3y
The constraints are 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
The feasible region is shaded in the figure. The point B is obtained by solving the equation 5x + 2y = 10 and 3x + 5y = 15.
Question 4.
Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solution:
The objective function is Z = 3x + 5y
The constraints are x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
The feasible region is shaded in the figure.
From the table the minimum value of Z is 7. Since the feasible region is unbounded, 7 may or may not be the minimum value of Z. Consider the graph of the inequality 3x + 5y < 7. This half plane has no point in common with the feasible region. Hence the minimum value of Z is 7 at B (\(\frac { 3 }{ 2 }\), \(\frac { 1 }{ 2 }\))
Question 5.
Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Solution:
The objective function is Z = 3x + 5y
The constraints are x + 2y < 10,
3x + y ≤ 15, x, y ≥ 0.
The feasible region is shaded in the figure. We use the corner point method to find the maximum of Z
Maximum value of Z is 18 at B(4, 3)
Question 6.
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Solution:
Consider 2x + y ≥ 3
Let 2x + y = 3
⇒ y = 3 – 2x
(0,0) is not contained in the required half plane as (0, 0) does not satisfy the in equation 2x + y ≥ 3.
Again consider x + 2y ≥ 6
Let x + 2y = 6
⇒ \(\frac { x }{ 6 } +\frac { y }{ 3 }\) = 1
Here also (0,0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its comers are A (6,0) and B (0,3). At A, Z = 6 + 0 = 6
At B, Z = 0 + 2 × 3 = 6
We see that at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z = 6 which is also minimum.
Question 7.
Minimise and Maximise Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0
Solution:
The objective function is Z = 5x + 10v.
The constraints are x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0
The feasible region is shaded in the figure. We use the comer point method to find the maximum/minimum value of Z.
Minimum value of Z is 300 at A(60,0).
The maximum value of Z is 600 at all points on the line joining B(120, 0) and C(60, 30).
Question 8.
Minimize and maximize Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
Solution:
The objective function is Z = x + 2y
The constraints are x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0
The feasible region is shaded in the figure. We use corner point method to find maxi-mum/minimum value of Z.
Maximum value of Z is 400 at B(0,200). The minimum value of Z is 100 at all points of the line joining the points (0,50) and (20,40).
Question 9.
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
Solution:
The objective function is Z = – x + 2y
The constraints are x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
The feasible region is shaded in the figure.
From the table the maximum value of Z is 1. Since the feasible region is unbounded 1 may or may not be the maximum value of Z
Consider the graph of the inequality – x + 2y > 1. This half plane has points common with the feasible region. Hence there is no maximum value for Z.
Question 10.
Maximize Z = x + y subject to x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0
Solution:
The objective function is Z = x + y
The constraints are x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0
There is no point satisfying the constraints simultaneously. Thus the problem has no feasible region. Hence no maximum value for Z.