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प्रश्न
A manufacturer produces bulbs and tubes. Each of these must be processed through two machines M1 and M2. A package of bulbs requires 1 hour of work on Machine M1 and 3 hours of work on M2. A package of tubes requires 2 hours on Machine M1 and 4 hours on Machine M2. He earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes. If maximum availability of Machine M1 is 10 hours and that of Machine M2 is 12 hours, then formulate the L.P.P. to maximize the profit.
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उत्तर
Let the manufacturer produce ‘x’ packages of bulbs and ‘y’ packages of tubes.
The profit on a package of bulbs is ₹ 13.5 and that of tubes is ₹ 55.
∴ Total profit = ₹ (13.5 x + 55y)
We construct a table with the constraints of machines M1 and M2 as follows:
| Machine\Product | Bulbs x |
Tubes y |
Maximum Availability in hours |
| M1 | 1 | 2 | 10 |
| M2 | 3 | 4 | 12 |
From the table, the total time required on M1 is (x + 2y) hours and on M2 is (3x + 4y) hours.
∴ The constraints are:
x + 2y ≤ 10, 3x + 4y ≤ 12
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0
∴ Given problem can be formulated as follows:
Maximize Z = 13.5x + 55y
Subject to x + 2y ≤ 10, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0.
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