Question

A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

Answer 1

Let the factory manufacture x screws of type A and y screws of type B on each day. Therefore,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as follows.

	Screw A	Screw B	Availability
Automatic Machine (min)	4	6	4 × 60 =240
Hand Operated Machine (min)	6	3	4 × 60 =240

The profit on a package of screws A is Rs 7 and on the package of screws B is Rs 10. Therefore, the constraints are

`4x+6y <= 240`

`6x + 3y < 240`

Total profit, Z = 7x + 10y

The mathematical formulation of the given problem is

Maximize Z = 7x + 10y … (1)

subject to the constraints,

`4x+6y <=240`… (2)

`6x +3y <= 240` … (3)

x, y ≥ 0 … (4)

The feasible region determined by the system of constraints is

The corner points are A (40, 0), B (30, 20), and C (0, 40).

The values of Z at these corner points are as follows.

Corner point	Z = 7x + 10y
A(40, 0)	280
B(30, 20)	410	→ Maximum
C(0, 40)	400

The maximum value of Z is 410 at (30, 20).

Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.