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.

Answer 1

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.