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प्रश्न
Solve the following L.P.P. graphically:
Maximise Z = x + 3y
subject to the constraints:
x + 2y ≤ 200
x + y ≤ 150
y ≤ 75
x, y ≥ 0
आलेख
उत्तर
Maximise Z = x + 3y
subject to the constraints:
x + 2y ≤ 200
x + y ≤ 150
y ≤ 75
x, y ≥ 0
Now,
| x | 0 | 200 | 100 | 50 |
| y | 100 | 0 | 50 | 75 |
x + y = 150
| x | 0 | 150 | 75 | 100 |
| y | 150 | 0 | 75 | 50 |
y = 75
| Corner point | z = x + 3y |
| (0, 0) | 0 + 3 (0) = 0 |
| (0, 75) | 0 + 3 × 75 = 225 |
| (50, 75) | 50 + 3 × 75 = 275 ⇒ (Maximise) |
| (100, 50) | 100 + 3 × 50 = 250 |
| (150, 0) | 150 + 3 × 0 = 150 |
Maximum value of z = 275 at x = 50, y = 75
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