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Question
Minimise Z = 13x – 15y subject to the constraints: x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0
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Solution
Given that: Z = 13x – 15y and the constraints
x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0
Let x + y = 7
| x | 3 | 4 |
| y | 4 | 3 |
Let 2x – 3y + 6 = 0
| x | 1 | -3 |
| y | 2 | 0 |
The shaded region is the feasible region determined by the constraints x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0
The feasible region is bounded with four corners O(0, 0), A(7, 0), B(3, 4), C(0, 2)
So, the maximum value can occur at any corner.
Let us evaluate the value of Z.
| Corner points | Value of Z | |
| O(0, 0) | 13(0) – 15(0) = 0 | |
| A(7, 0) | 13(7) – 15(0) = 91 | |
| B(3, 4) | 13(3) – 15(4) = – 21 | |
| C(0, 2) | 13(0) – 15(2) = – 30 | ← Minimum |
Hence, the minimum value of Z is – 30 at (0, 2).
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