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Question
Solve the following LP.P.
Maximize z = 13x + 9y,
Subject to 3x + 2y ≤ 12,
x + y ≥ 4,
x ≥ 0,
y ≥ 0.
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Solution
| Equation | x | y | Points | Side |
| 3x + 2y = 12 | 0 | 6 | A(0, 6) | Origin |
| 4 | 0 | B(4, 0) | Side | |
| x + y = 4 | 0 | 4 | C(0, 4) | Non-origin |
| 4 | 0 | D(4, 0) | Side |
Shaded region is the feasible region ABCA.
Z = 13x + 9y
Z(A) = 0 + 9 × 6 = 54
Z(B) = 13 × 4 + 0 = 52
Z(C) = 0 + 4 × 9 = 36
∴ Max, value of z is 54 at A(0, 6)
i.e., when x = 0,
y = 6.
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