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Question
Determine the maximum value of Z = 3x + 4y if the feasible region (shaded) for a LPP is shown in Figure
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Solution
OAED is the feasible region.
At A, y = 0
∴ 2x + y = 104
⇒ x = 52
At D, x = 0
∴ x + 2y = 76
⇒ y = 38
Which gives corner point D = (0, 38)
Now solving the given equations, we get
x + 2y = 76
2x + y = 104
2x + 4y = 152
2x + y = 104
(–) (–) (–)
3y = 48
⇒ y = 16
x + 2(16) = 76
⇒ x = 76 – 32 = 44
So, the corner point E = (44, 16)
Evaluating the maximum value of Z, we get
| Corner points | Z = 3x + 4y | |
| O(0, 0) | Z = 3(0) + 4(0) = 0 | |
| A(52, 0) | Z = 3(52) + 4(0) = 156 | |
| E(44, 16) | Z = 3(44) + 4(16) = 196 | ← Maximum |
| D(0, 38) | Z = 3(0) + 4(38) = 152 |
Hence, the maximum value of Z is 196 at (44, 16).
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