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Question
Maximize Z = 400x + 500y subject to constraints
x + 2y ≤ 80, 2x + y ≤ 90, x ≥ 0, y ≥ 0
Solution
To draw the feasible region, construct table as follows:
| Inequality | x + 2y ≤ 80 | 2x + y ≤ 90 |
| Corresponding equation (of line) | x + 2y = 80 | 2x + y = 90 |
| Intersection of line with X-axis | (80, 0) | (45, 0) |
| Intersection of line with Y-axis | (0, 40) | (0, 90) |
| Region | Origin side | Origin side |
Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(45, 0), B and C(0, 40).
B is the point of intersection of the lines 2x + y = 90 and x + 2y = 80.
Solving the above equations, we get
x = `100/3`, y = `70/3`
∴ B = `(100/3, 70/3)`
Here, the objective function is
Z = 400x + 500y
∴ Z at O(0, 0) = 400(0) + 500(0) = 0
Z at A(45, 0) = 400(45) + 500(0) = 18000
X at B`(100/3, 70/3) = 400(100/3) + 500(70/3)`
= `40000/3 + 35000/3`
= `75000/3`
= 25000
Z at C(0, 40) = 400(0) + 500(40)
= 20000
∴ Z has maximum value 25000 at x = `100/3` and y = `70/3`.
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