Solve the following L.P.P. graphically:
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0 and x, y ≥ 0
Advertisement Remove all ads
Solution
The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x ≥ 0, and y ≥ 0, is as follows.
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.
Corner point | Z = 5x + 10y | |
A(60, 0) | 300 | →Minimum |
B(120, 0) | 600 | →Maximum |
C(60, 30) | 600 | →Maximum |
D(40, 20) | 400 |
The minimum value of Z is 300 at (60, 0).
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads