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Solve the linear programming problem 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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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 co-ordinates of the comer points are A(12, 0), B(4, 2), C(2, 6) and D(0, 18).

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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