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in Operations Research by (26.1k points)
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Solve the linear programming problem by graphical method.

Maximize Z = 6x1 + 8x2 subject to constraints 30x1 + 20x2 ≤ 300; 5x1 + 10x2 ≤ 110; and x1, x2 ≥ 0.

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Best answer

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 eo-ordinates of B)

3x1 + 2x2 = 30 … (1)

x1 + 2x2 = 22 … (2)

(1) – (2) ⇒ 2x1 = 8

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).

The maximum value of Z occurs at B.

∴ The optimal solution is x1 = 4, x2 = 9 and Zmax = 96.

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