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Question
Maximize z = 5x + 2y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Solution
To draw the feasible region, construct table as follows:
| Inequality | 3x + 5y ≤ 15 | 5x + 2y ≤ 10 |
| Corresponding equation (of line) | 3x + 5y = 15 | 5x + 2y = 10 |
| Intersection of line with X-axis | (5, 0) | (2, 0) |
| Intersection of line with Y-axis | (0, 3) | (0, 5) |
| Region | Origin side | Origin side |
x ≥ 0, y ≥ 0 represent 1st quadrant.
Here, the objective function is Z = 5x + 2y
∴ Z at O(0, 0) = 5(0) + 2(0) = 0
Z at Q(2, 0) = 5(2) + 2(0) = 10
Z at R `(20/19, 45/19) = 5 (20/19) + 2(45/19) = 10`
Z at A(0, 3) = 5(0) + 2(3) = 6
The maximum value of Z is 10 and it occurs at every point lying on the line segment joining `"R"(20/19, 45/19)` and Q(2, 0).
Hence, there are infinitely many optimal solutions.
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