Question
Solve the linear programming problem and determine the maximum profit to the manufacturer.
✓
Answer
We have Maximise Z = 100x + 170y Subject to
$3\text{x}+2\text{y}\leq3600,\text{x}+4\text{y}\leq1800,\text{x}\geq0,\text{y}\geq0$
From the shaded feasible region it is clear that the coordinates of corner points are (0, 0), (1200, 0), (1080, 180) and (0, 450).
On solving x + 4y = 1800 and 3x + 2y = 3600, we get x = 1080 and y = 180.
Hence, the maximum profit to the manufacture is 138600.
$3\text{x}+2\text{y}\leq3600,\text{x}+4\text{y}\leq1800,\text{x}\geq0,\text{y}\geq0$
From the shaded feasible region it is clear that the coordinates of corner points are (0, 0), (1200, 0), (1080, 180) and (0, 450).
On solving x + 4y = 1800 and 3x + 2y = 3600, we get x = 1080 and y = 180.
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Corner points
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Corresponding value of Z = 100x + 170y
|
|
(0, 0)
(1200, 0)
(1080, 180)
(0, 450)
|
0
1200 ×100 = 12000
100 × 1080 + 170 × 180 = 138600 (maximum)
0 + 170 × 450 = 76500
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