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Solve the linear programming problem by graphical method.

Maximize Z = 40x1 + 50x2 subject to constraints 3x1 + x2 ≤ 9; x1 + 2x2 ≤ 8 and x1, x2 ≥ 0.

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Given that 3x1 + x2 ≤ 9

Let 3x1 + x2 = 9

x1 0 3
x2 9 0

Also given that x1 + 2x2 ≤ 8

Let x1 + 2x2 = 8

x1 0 8
x2 4 0

3x1 + x2 = 9 … (1)

x1 + 2x2 = 8 … (2)

(1) x 2 ⇒ 6x1 + 2x2 = 18 ... (3)

(2) + (3) ⇒ -5x1 = -10

x1 = 2

x1 = 2 substitute in (1)

3(2) + x2 = 9

x2 = 3

The feasible region satisfying all the conditions is OABC.

The co-ordinates of the corner points are O(0, 0), A(3, 0), B(2, 3), C(0, 4)

The maximum value of Z occurs at (2, 3).

∴ The optimal solution is x1 = 2, x2 = 3 and Zmax = 230

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