A factory manufactures two type 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 Rs. 10. 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 maximise his profit ? Determine the maximum profit.

Let the manufacturer produce x packages of screw A and y packages of screw B.
Let Z be the profit. 
Table

We are to maximise
P = 7x + 10y
subject to constraints
4x + 6y ≤ 240  or  2x + 3y ≤ 120
6x + 3y ≤ 240 or 2x + y ≤ 80
x ≥ 0. y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
We draw the graph of 2x + 3y = 120
For a = 0, 3y = 120 or y = 40
For y = 0, 2x = 120 or x = 60
∴ line meets OX in A(60, 0) and OY in L(0, 40)
Again we draw the graph of 2x + y = 80
For x = 0, y = 80
For y = 0, 2 x = 80 or x = 40
∴ line meets OX in B(40, 0) and OY in M(0, 80),
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.

The corner points are O(0, 0), B(40, 0), C(30, 20), L(0, 40)
At O(0, 0), P = 7 × 0 + 10 × 0 = 0 + 0 = 0
At B(40, 0), P = 7 × 40 + 10 × 0 = 280 + 0 = 280
At C(30, 20), P = 7 × 30 + 10 × 20 = 210 + 200 = 410
At L(0, 40), P = 7 × 0 + 10 × 40 = 0 + 400 = 400
∴  maximum value = 410 at (30, 20)
∴ 30 packages of screws A and 20 packages of screws B arc produced for maximum profit of Rs. 410.