A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture m order to maximise the profit?

Explanation

Let `x` be the souvenirs of A type and `y` of B type be manufacture by the company. Then,

Maximise `Z=5x+6y`……………1
Constraints `5x+8yle200`………………2
`10x+8yle240implies4x+4yle120`………….3
`xge0,yge0`.................4
First, draw the graph of first line `5x+8y=200`.

Put `(0,0)` in the inequation `5x+8yle200`,
`5xx0+8xx0le200implies0le200`. (True)
Therefore the half plane contains the origin.
Since `x,yge0`.So the feasible region will be in first quadrant.
Now, draw the graph of the line `5x+4y=120`.

Put `(0,0)` in the inequation `5x+4yle120`,
`5xx0+4xx0le120`
`implies0le120` (True)
Therefore, the half plane contains the origin.
From equations `5x+8y=200` and `5x+4y=120`.
The point of intersection is `B(8,20)`.
`:.` Feasible region is OABCEO.
Thus, the vertices of the feasible region are `O(0,0),A(24,0),B(8,20)` and `C(0,25)`. We find the value of `Z` at these vertices .

The maximum value of `Z` is Rs. 160 at point `B(8,20)`.Therefore, to obtain the maximum profit Rs. 160 the souvenirs 8 of types of A and 20 of tyes B should be produced.