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Question
Minimize Z = x + 4y subject to constraints
x + 3y ≥ 3, 2x + y ≥ 2, x ≥ 0, y ≥ 0
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Solution
To draw the feasible region, construct table as follows:
| Inequality | x + 3y ≥ 3 | 2x + y ≥ 2 |
| Corresponding equation (of line) | x + 3y = 3 | 2x + y = 2 |
| Intersection of line with X-axis | (3, 0) | (1, 0) |
| Intersection of line with Y-axis | (0, 1) | (0, 2) |
| Region | Non-origin side | Non-origin side |
Shaded portion XABCY is the feasible region, whose vertices are A(3, 0), B and C(0, 2).
B is the point of intersection of the lines 2x + y = 2 and x + 3y = 3.
∴ B ≡ `(3/5, 4/5)`
Here, the objective function is
Z = x + 4y
∴ Z at A(3, 0) = 3 + 4(0)
= 3
Z at B`(3/5, 4/5) = 3/5 + 4(4/5)`
= `19/5`
= 3.8
Z at C(0, 2) = 0 + 4(2)
= 8
∴ Z has minimum value 3 at x = 3 and y = 0.
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