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प्रश्न
Solve the following Linear Programming problem graphically:
Maximie Z = 300x + 600y
Subject to x + 2y ≤ 12
2x + y ≤ 12
x + `5/4`y ≥ 5
x ≤ 0, y ≥ 0.
योग
उत्तर
Since Z = 300x + 600y
x + 2y ≤ 12
2x + y ≤ 12
x + `5/4`y ≥ 5
x ≤ 0, y ≥ 0
Now, x + 2y = 12 2x + y = 12
| x | 0 | 12 | 4 |
| y | 6 | 0 | 4 |
| x | 0 | 6 | 4 |
| y | 12 | 0 | 4 |
`x + 5/4 y = 5`
| x | 0 | 5 |
| y | 4 | 0 |
| Corner point | z = 300x + 600y |
| A(5, 0) | z = 1500 |
| B(6, 0) | z = 1800 |
| C(4, 4) | z = 3600 Maximum |
| D(0, 6) | z = 3600 Maximum |
| E(0, 4) | z = 2400 |
The maximum of objective function at two points at (4, 4) and (0, 6).
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