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Question
Maximize Z = 2x + 3y subject to constraints
x + 4y ≤ 8, 3x + 2y ≤ 14, x ≥ 0, y ≥ 0.
Solution
To draw the feasible region, construct table as follows:
| Inequality | x + 4y ≤ 8 | 3x + 2y ≤ 14 |
| Corresponding equation (of line | x + 4y = 8 | 3x + 2y = 14 |
| Intersection of line with X-axis | (8, 0) | `(14/3, 0)` |
| Intersection of line with Y-axis | (0, 2) | (0, 7) |
| Region | Origin side | Origin side |
Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A`(14/3, 0)`, B and C(0, 2).
B is the point of intersection of the lines x + 4y = 8 and 3x + 2y = 14.
Solving the above equations, we get
x = 4, y = 1
∴ B = (4, 1)
Here, the objective function is
Z = 2x + 3y
∴ Z at O(0, 0) = 2(0) + 3(0) = 0
Z at A`(14/3, 0) = 2(14/3) + 3(0) = 28/3`
Z at B(4, 1) = 2(4) + 3(1) = 8 + 3 = 11
Z at C(0, 2) = 2(0) + 3(2) = 6
∴ Z has maximum value 11 at x = 4 and y = 1.
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