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Question
Solve the following linear programming problems by graphical method.
Maximize Z = 6x1 + 8x2 subject to constraints 30x1 + 20x2 ≤ 300; 5x1 + 10x2 ≤ 110; and x1, x2 ≥ 0.
Solution
Given that 30x1 + 20x2 ≤ 300
Let 30x1 + 20x2 = 300
Therefore 3x1 + 2x2 = 30
| x1 | 0 | 10 |
| x2 | 15 | 0 |
Also given that 5x1 + 10x2 ≤ 110
Let 5x1 + 10x2 = 110
x1 + 2x2 = 22
| x1 | 0 | 22 |
| x2 | 11 | 0 |
To get point of intersection, (i.e., the to get co-ordinates of B)
3x1 + 2x2 = 30 …….(1)
x1 + 2x2 = 22 ……..(2)
2x1 = 8 ......[Equation (1) – (2)]
x1 = 4
x1 = 4 substitute in (1),
x1 + 2x2 = 22
4 + 2x2 = 22
2x2 = 18
x2 = 9
i.e., B is (4, 9)
The feasible region satisfying all the given conditions is OABC.
The co-ordinates of the points are O(0, 0), A(10, 0), B(4, 9), C(0, 11).
| Corner points | Z = 6x1 + 8x2 |
| O(0, 0) | 0 |
| A(10, 0) | 60 |
| B(4, 9) | 6 × 4 + 8 × 9 = 96 |
| C(0, 11) | 88 |
The maximum value of Z occurs at B(4, 9).
∴ The optimal solution is x1 = 4, x2 = 9 and Zmax = 96
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