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Question
Minimize z = 7x + y subjected to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
Solution
To draw the feasible region, construct table as follows:
| Inequality | 5x + y ≥ 5 | x + y ≥ 3 |
| Corresponding equation (of line) | 5x + y = 5 | x + y = 3 |
| Intersection of line with X-axis | (1, 0) | (3, 0) |
| Intersection of line with Y-axis | (0, 5) | (0, 3) |
| Region | Non-origin side | Non-origin side |
x ≥ 0, y ≥ 0 represent 1st quadrant.
Shaded portion XABCY is the feasible region, whose vertices are A(3, 0), B and C (0, 5).
B is the point of intersection of the lines x + y = 3 and 5x + y = 5.
Solving the above equations, we get x = `1/2`, y = `5/2`
∴ B ≡ `(1/2, 5/2)`
Here, the objective function is Z = 7x + y
∴ Z at A(3, 0) = 7(3) + 0 = 21
Z at B`(1/2, 5/2) = 7(1/2) + 5/2 = 7/2 + 5/2` = 6
Z at C (0, 5) = 7(0) + 5 = 5
∴ Z has minimum value 5 at C(0, 5).
∴ Z has minimum value 5 when x = 0 and y = 5.
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