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Question
Minimize Z = 2x + 3y subject to constraints
x + y ≥ 6, 2x + y ≥ 7, x + 4y ≥ 8, x ≥ 0, y ≥ 0
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Solution
To draw the feasible region, construct table as follows:
| Inequality | x + y ≥ 6 | 2x + y ≥ 7 | x + 4y ≥ 8 |
| Corresponding equation (of line) | x + y = 6 | 2x + y = 7 | x + 4y = 8 |
| Intersection of line with X-axis | (6, 0) | `(7/2, 0)` | (8, 0) |
| Intersection of line with Y-axis | (0, 6) | (0, 7) | (0, 2) |
| Region | Non-origin side | Non-origin side | Non-origin side |
Shaded portion XABCDY is the feasible region, whose vertices are A(8, 0), B, C and D(0, 7).
B is the point of intersection of x + 4y = 8 and x + y = 6.
Solving the above equations, we get
x = `16/3`, y = `2/3`
∴ B ≡ `(16/3, 2/3)`
C is the point of intersection of 2x + y = 7 and x + y = 6.
Solving the above equations, we get
x = 1, y = 5
∴ C ≡ (1, 5)
Here, the objective function is Z = 2x + 3y
∴ Z at A(8, 0) = 2(8) + 3(0) = 16
Z at B`(16/3, 2/3)`
= `2(16/3) + 3(2/3)`
= `32/3 + 6/3`
= `38/3`
Z at C(1, 5) = 2(1) + 3(5) = 17
Z at D(0, 7) = 2(0) + 3(7) = 21
∴ Z has minimum value `38/3` at x = `16/3` and y = `2/3`.
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