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Question
Maximize z = 7x + 11y subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0
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Solution
To draw the feasible region, construct table as follows:
| Inequality | x + 5y ≤ 26 | 5x + 3y ≤ 30 |
| Corresponding equation (of line) | 3x + 5y = 26 | 5x + 3y = 30 |
| Intersection of line with X-axis | `(26/3, 0)` | (6, 0) |
| Intersection of line with Y-axis | `(0, 26/5)` | (0, 10) |
| Region | Origin side | Origin side |
x ≥ 0, y ≥ 0 represent 1st quadrant.
Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(6, 0), B and C `(0, 26/5)`.
B is the point of intersection of the lines 5x + 3y = 30 and 3x + 5y = 26. Solving the above equations, we get
x = `9/2`, y = `5/2`
∴ B ≡ `(9/2, 5/2)` ≡ `(4.5, 2.5)`
Here, the objective function is Z = 7x + 11y
∴ Z at O(0, 0) = 7(0) + 11(0) = 0
Z at A(6, 0) = 7(6) + 11(0) = 42
Z at B`(9/2, 5/2) = 7(9/2) + 11(5/2)`
= `(63 + 55)/2`
= 59
Z at C`(0, 26/5) - 7(0) + 11(26/5)`
= `286/5`
= 57.2
∴ Z has maximum value 59 at B`(9/2, 5/2)`.
i.e. Z has maximum value 59 when x = 4.5 and y = 2.5
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