A factory manufactures two types of screws, A and B, each type 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 Rs. 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produced in a day in order to maximize his profit? Determine the maximum profit.

Let the manufacturer produces x packages of screws a and y packages of screws B, then time taken by x packages of screws A and y packages of screws B on automatic machine=(4x+6y) minutes. And hand operated machine = (6x+3y) minutes. 

As, each machine is available for at the most 4 hours i.e., 

4 × 60 = 240 minutes. 

So, we have 4x + 6y ≤ 240 i.e., 

2x + 3y ≤ 120 and 6x + 3y ≤ 240 i.e., 2x + y ≤ 80

Profit on selling x packages of screws A and y packages of screws B is Z = 7x + 10y

So, our problem is to find x and y such that Z = 7x + 10y is maximum subject to 

2x + 3y ≤ 120, 2x + y ≤ 80, x ≥ 0"and" y ≥ 0.

Plot the graphs of equations 2x + 3y = 120 "and" 2x + y = 80

The feasible portion of the graph satisfying the inequalities 2x + 3y ≤ 120 and 2x + y ≤ 80 is OABC which is shown shaded in the figure.

Coordinates of O, A, B and C are (0, 0) (0, 40) (30, 20) and (40, 0).

We will find the values of Z at corner points O, A, B and C 

At A(0, 40),       Z = 7x + 10y = 0 + 10 × 40 = 400            

At B(30, 200     Z = 7 × 30 + 10 × 20 = 210 + 200 = 410 

At C(40, 0)       Z = 7 × 40 + 0 = 280  

At O(0, 0)          Z = 0

Hence maximum profit Rs 410 when 30 screws of type A and 20 screws of type B are produced.