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 on screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacturer 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 order to maximise his profit? Determine the maximum profit.

Explanation

Let the manufacturer makes `x` packets of screw `A` and `y` packets of screw B per day. Then

Maximise `Z=7x+10y`………………1
and constraints
`4x+6yle240implies2x+3yle120`…………………2
`6x+3yle240implies2x+yle80`.....................3
and `xge0,yge0`

First draw the graph of the line `2x+3y=120`

Put `(0,0)` in the inequation `2x+3yle120`
`2xx0+3xx0le120`
`implies 0le120` (True)
Therefore, half plane contains the origin,
Now draw the graph of the line `2x+y=80`

Put `(0,0)` in the inequation `2x+yle80`
`2xx0+0le80implies0le80` (True)
Therefore half plane contains the origin.Since, `x,yge0`. So the feasible region will be in first quadrant.
From equations `2x+3y=120` and `2x+y=80` the point of intersection is `B(30,20)`.
`:.` Feasible region is OABCE.
The vertices of the feasible region are `O(0,0), A(40,0),B(30,20)` and `C(0,40)` we find the value of `Z` at these points.

Therefore, the maximum value of `Z` is Rs. 410 at point `B(30,20)`. Therefore, the maximum profit of Rs. 410 will be obtained when the factory produces 30 packets of screw A and 20 packets of screw B per day.