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Question
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.
Solution
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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