Question

A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B per unit and the number of man-hours available for the firm is as follows:

Gadgets	Foundry	Machine shop
A	10	5
B	6	4
Time available (hour)	60	35

Profit on the sale of A is ₹ 30 and B is ₹ 20 per units. Formulate the L.P.P. to have maximum profit.

Answer 1

Let the number of gadgets A produced by the firm be x and the number of gadgets B produced by the firm be y.

The profit on the sale of A is ₹ 30 per unit and on the sale of B is ₹ 20 per unit.

∴ Total profit is z = 30x + 20y

This is a linear function that is to be maximized. Hence, it is the objective function. The constraints are as per the following table:

	Gadgets A (x)	Gadgets B (y)	Total available Time (in hour)
Foundry	10	6	60
Machine shop	5	4	35

From the table, the total man-hours of labour required for x units of gadget A and y units of gadget B in the foundry is (10x + 6y) hours, and the total man-hours of labour required in a machine shop is (5x + 4y) hours.

Since the maximum time available in foundry and machine shops are 60 hours and 35 hours respectively.

Therefore, the constraints are 10x + 6y ≤ 60, 5x + 4y ≤ 35.

Since x and y cannot be negative, we have x ≥ 0, y ≥ 0.  Hence, the given LPP can be formulated as:

Maximize z = 30x + 20y, subject to

10x + 6y ≤ 60,

5x + 4y ≤ 35,

x ≥ 0, y ≥ 0