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Question
Solve the following linear programming problems by graphical method.
Minimize Z = 20x1 + 40x2 subject to the constraints 36x1 + 6x2 ≥ 108; 3x1 + 12x2 ≥ 36; 20x1 + 10x2 ≥ 100 and x1, x2 ≥ 0.
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Solution
Given that 36x1 + 6x2 ≥ 108
Let 36x1 + 6x2 = 108
6x1 + x2 = 18
| x1 | 0 | 3 | 2 |
| x2 | 18 | 0 | 6 |
Also given that 3x1 + 12x2 ≥ 36
Let 3x1 + 12x2 = 36
x1 + 4x2 = 12
| x1 | 0 | 12 | 4 |
| x2 | 3 | 0 | 2 |
Also given that 20x1 + 10x2 ≥ 100
Let 20x1 + 10x2 = 100
2x1 + x2 = 10
| x1 | 0 | 5 | 4 |
| x2 | 10 | 0 | 2 |
The feasible region satisfying all the conditions is ABCD.
The coordinates of the comer points are A(12, 0), B(4, 2), C(2, 6) and D(0, 18).
| Corner points | Z = 20x1 + 40x2 |
| A(12, 0) | 240 |
| B(4, 2) | 80 + 80 = 160 |
| C(2, 6) | 40 + 240 = 280 |
| D(0, 18) | 720 |
The minimum value of Z occurs at B(4, 2)
∴ The optimal solution is x1 = 4, x2 = 2 and Zmin = 160
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