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Question
Solve the following linear programming problems by graphical method.
Maximize Z = 22x1 + 18x2 subject to constraints 960x1 + 640x2 ≤ 15360; x1 + x2 ≤ 20 and x1, x2 ≥ 0.
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Solution
Given that 960x1 + 640x2 ≤ 15360
Let 960x1 + 640x2 = 15360
3x1 + 2x2 = 48
| x1 | 0 | 16 |
| x2 | 24 | 0 |
Also given that x1 + x2 ≤ 20
Let x1 + x2 = 20
| x1 | 0 | 20 |
| x2 | 20 | 0 |
To get point of intersection
3x1 + 2x2 = 48 …..(1)
x1 + x2 = 20 ……(2)
− 2x1 – 2x2 = – 40 …..(3) ......[Equation (2) × –2]
x1 = 8 .....[Adding equation (1) and (3)]
x1 = 8 substitute in (2),
8 + x2 = 20
x2 = 12
The feasible region satisfying all the given conditions is OABC.
The co-ordinates of the comer points are O(0, 0), A(16, 0), B(8,12) and C(0, 16).
| Corner points | Z = 22x1 + 18x2 |
| O(0, 0) | 0 |
| A(16, 0) | 352 |
| B(8, 12) | 392 |
| C(0, 20) | 360 |
The maximum value of Z occurs at B(8, 12).
∴ The optimal solution is x1 = 8, x2 = 12 and Zmax = 392
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