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Question
Solve the following L.P.P. by graphical method:
Maximize: Z = 4x + 6y
Subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
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Solution
The draw the feasible region, construct table as follows:
| Inequality | 3x + 2y ≤ 12 | x + y ≥ 4 |
| Corresponding equation (of line) | 3x + 2y = 12 | x + y = 4 |
| Intersection of line with X-axis | (4, 0) | (4, 0) |
| Intersection of line with Y-axis | (0, 6) | (0, 4) |
| Region | Origin side | Non-origin side |
Shaded portion ABC is the feasible region,
Whose vertices are A(4, 0), B(0, 6), C(0, 4).
Here, the objective function is Z = 4x + 6y
∴ Z at A(4, 0) = 4(4) + 6(0) = 16
Z at B(0, 6) = 4(0) + 6(6) = 36
Z at C(0, 4) = 4(0) + 6(4) = 24
∴ Z has maximum value 36 at B(0, 6)
∴ Z is maximum when x = 0 and y = 6.
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