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Maximize Z = 5x + 10y subject to constraints
x + 2y ≤ 10, 3x + y ≤ 12, x ≥ 0, y ≥ 0
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Solution
To draw the feasible region, construct table as follows:
Inequality | x + 2y ≤ 10 | 3x + y ≤ 12 |
Corresponding equation (of line) | x + 2y = 10 | 3x + y = 12 |
Intersection of line with X-axis | (10, 0) | (4, 0) |
Intersection of line with Y-axis | (0, 5) | (0, 12) |
Region | Origin side | Origin side |
Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(4, 0), B and C(0, 5).
B is the point of intersection of the lines x + 2y = 10 and 3x + y = 12.
Solving the above equations, we get
x = `14/5` y = `18/5`
Here, the objective function is
Z = 5x + 10y
∴ Z at O(0, 0) = 5(0) + 10(0)
= 0
Z at A(4, 0) = 5(4) + 10(0)
= 20
Z at B`(14/5, 18/5) = 5(14/5) + 10(18/5)`
= 14 + 36
= 50
Z at C(0, 5) = 5(0) + 10(5)
= 50
The maximum value of Z is 50 and it occurs at every point lying on the line segment joining B`(14/5, 18/5)` and C(0, 5).
Hence, there are infinitely many optimal solutions.
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