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Question
Solve the following linear programming problems by graphical method.
Maximize Z = 20x1 + 30x2 subject to constraints 3x1 + 3x2 ≤ 36; 5x1 + 2x2 ≤ 50; 2x1 + 6x2 ≤ 60 and x1, x2 ≥ 0.
Solution
Given that 3x1 + 3x2 ≤ 36
Let 3x1 + 3x2 = 36
| x1 | 0 | 12 |
| x2 | 12 | 0 |
Also given that 5x1 + 2x2 ≤ 50
Let 5x1 + 2x2 = 50
| x1 | 0 | 10 |
| x2 | 25 | 0 |
3x1 + 3x2 = 36
x1 + x2 = 12 ……….(1)
5x1 + 2x2 = 50 ………(2)
2x1 + 2x2 = 24 ....[(1) × 2]
− − −
−3x1 = − 6
x1 = 2
Substituting x1 = 2 in (1) we get
2+ x2 = 12
x2 = 6
Also given that 2x1 + 6x2 ≤ 60
Let 2x1 + 6x2 = 60
x1 + 3x2 = 30
| x1 | 0 | 30 |
| x2 | 10 | 0 |
x1 + x2 = 12 …….(1)
x1 + 3x2 = 30 …….(2)
– 2x2 = – 18 ......[Equation (1) – (2)]
x2 = 9
x2 = 9 substitute in (1)
x1 + x2 = 12
x1 + 9 = 12
x1 = 12 – 9
x1 = 3
The feasible region satisfying all the given conditions is OABCD.
The co-ordinates of the comer points are
| Corner points | Z = 20x1 + 30x2 |
| O(0, 0) | 0 |
| A(10, 0) | 200 |
| B(2, 6) | 220 |
| C(3, 9) | 330 |
| D(0, 10) | 300 |
The maximum value of Z occurs at C(3, 9)
∴ The optimal solution is x1 = 3, x2 = 9 and Zmax = 330
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