A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand - operated. It takes 4 minute on the automatic and 6 minutes on the hand - operated machines to manufacture a package of screws ‘A’, while it takes 6 minutes on the automatic and 3 minutes on the hand - operated machine to manufacture a package of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws ‘A’ at a profit of 70 P and screws ‘B’ at a profit of ₹ 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

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

4x + 6y 240

6x + 3y 240

The manufacturer can sell a package of screws ‘A’ at a profit of Rs 0.7 and screws ‘B’ at a profit of Re 1.

Total profit, Z = 0.7x + 1y

The mathematical formulation of the given problem is

Maximize Z = 0.7x + 1y

subject to the constraints,

4x + 6y 240

6x + 3y 240

x, y 0

First we will convert the inequations into equations as follows:

4x + 6y = 240, 6x + 3y = 240, x = 0, y = 0.

Region represented by 4x + 6y 240

The line 4x + 6y = 240 meets the coordinate axes at A(60,0) and B(0,40) respectively. By joining these points we obtain the line 4x + 6y = 240. Clearly (0, 0) satisfies the 4x + 6y 240. So, the region which contains the origin represents the solution set of the inequation 4x + 6y 240.

Region represented by 6x + 3y 240

The line 6x + 3y = 240 meets the coordinate axes at C(40,0) and d(0,80) respectively. By joining these points we obtain the line 6x + 3y = 240. Clearly (0, 0) satisfies the 6x + 3y 240. So, the region which contains the origin represents the solution set of the inequation 6x + 3y 240.

Region represented by x 0, y 0 :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 0 and y 0.

The feasible region determined by the system of constraints 4x + 6y 240, 6x + 3y 240, x 0,

y 0 are as follows.

The corner points are C(40,0), E(30,20), B(0,40). The values of Z at these corner points are as follows

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.