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.

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Solution

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 =120

Hand Operated Machine (min)

6

3

4 × 60 =120

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

Total profit, Z = 7x + 10y

The mathematical formulation of the given problem is

Maximize Z = 7x + 10y … (1)

subject to the constraints,

… (2)

… (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.

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