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Question
Solve the following LPP by graphical method:
Maximize z = 11x + 8y, subject to x ≤ 4, y ≤ 6, x + y ≤ 6, x ≥ 0, y ≥ 0
Solution
To draw the feasible region, construct table as follows:
| Inequality | x ≤ 4 | y ≤ 6 | x + y ≤ 6 |
| Corresponding equation (of line) | x = 4 | y = 6 | x + y = 6 |
| Intersection of line with X-axis | (4, 0) | – | (6, 0) |
| Intersection of line with Y-axis | – | (0, 6) | (0, 6) |
| Region | Origin side | Origin side | Origin side |
Shaded portion OABC is the feasible region,
whose vertices are O(0, 0), A(4, 0), B and C(0, 6).
B is the point of intersection of the lines x = 4 and x + y = 6
Substituting x = 4 in x + y = 6, we get
y = 2
∴ B ≡ (4, 2)
Here, the objective function is Z = 11x + 8y
∴ Z at O(0, 0) = 11(0) + 8(0) = 0
Z at A(4, 0) = 11(4) + 8(0) = 44
Z at B(4, 2) = 11(4) + 8(2) = 44 + 16 = 60
Z at C(0, 6) = 11(0) + 8(6) = 48
∴ Z has maximum value 60 at B(4, 2)
∴ Z is maximum, when x = 4 and y = 2.
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In the figure, ABCD represents
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| A | (12, 0) | 4(12) + 5(0) | 48 | |
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| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
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