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Question
Solve the following linear programming problems by graphical method.
Minimize Z = 3x1 + 2x2 subject to the constraints 5x1 + x2 ≥ 10; x1 + x2 ≥ 6; x1 + 4x2 ≥ 12 and x1, x2 ≥ 0.
Solution
Given that 5x1 + x2 ≥ 10
Let 5x1 + x2 = 10
| x1 | 0 | 2 |
| x2 | 10 | 0 |
Also given that x1 + x2 ≥ 6
Let x1 + x2 = 6
| x1 | 0 | 6 |
| x2 | 6 | 0 |
Also given that x1 + 4x2 ≥ 12
Let x1 + 4x2 = 12
| x1 | 0 | 12 |
| x2 | 3 | 0 |
To get B
5x1 + x2 = 10 ……..(1)
x1 + x2 = 6 ………(2)
4x1 = 4 ......[Equation (1) – (2)]
x1 = 1
x = 1 substitute in (2)
x1 + x2 = 6
1 + x2 = 6
x2 = 5
∴ B is (1, 5)
To get C
x1 + x2 = 6
x1 + 4x2 = 12
− 3x2 = − 6 ..........[Equation (1) – (2)]
x2 = 2
x2 = 2 substitute in (2) we get,
x1 + x2 = 6
x1 = 4
∴ C is (4, 2)
The feasible region satisfying all the conditions is ABCD.
The coordinates of the comer points are A(0, 10), B(1, 5), C(4, 2) and D(12, 0).
| Corner points | Z = 3x1 + 2x2 |
| A(0, 10) | 20 |
| B(1, 5) | 13 |
| C(4, 2) | 16 |
| D(12, 0) | 36 |
The minimum value of Z occours at B(1, 5).
∴ The optimal solution is x1 = 1, x2 = 5 and Zmin = 13
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