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Question
Minimise Z = 3x + 5y subject to the constraints:
x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0
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Solution
We first draw the graphs of x + 2y = 10
x + y = 6
3x + y = 8.
The shaded region ABCD is the feasible region (R) determined by the above constraints.
The feasible region is unbounded.
Therefore, minimum of Z may or may not occur.
If it occurs, it will be on the corner point.
| Corner Point | Value of Z | |
| A (0, 8) | 40 | |
| B (1, 5) | 28 | |
| C (2, 4) | 26 | ← Smallest |
| D (10, 0) | 30 |
Let us draw the graph of 3x + 5y < 26 as shown in Figure by dotted line.
We see that the open half-plane determined by 3x + 5y < 26 and R do not have a point in common.
Thus, 26 is the minimum value of Z.
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