Maximise Z = 5x + 3y subject to 3x+5y≤15, 5x+2y≤10,x≥0,y≥0.

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asked Jan 25, 2018 in Mathematics by sforrest072 (157,439 points) 61 410 947

Maximise Z = 5x + 3y
subject to 3x+5y≤15, 5x+2y≤10,x≥0,y≥0.

1 Answer

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answered Jan 25, 2018 by mdsamim (213,225 points) 5 10 15
edited Mar 6, 2018 by faiz
 
Best answer

The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.​

The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3), and C(20/19,45/19)
The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is 235/19 at the point (20/19,45/19)

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