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A cottage industry manufactures pedestal...

A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs. 25 and that from a shade is Rs. 15. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit. Formulate an LPP and solve it graphically.

Text Solution

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Let the manufacture makes `x` pedestal lamps and `y` wooden shades per day then

Maximise `Z=5x+3y`…………. 1
Constraints `2x+yle12`…………………2
`3x+2yle20`………………..3
`xge0,yge0`………………….4
First, draw the graph of the equation `2x+y=12`

Put `(0,0)` in the inequation `2x+yle12`,
`2xx0+0le12`
`0le12` (True)
Therefore, half plane contain the origin.
Since `x,yge0`
Therefore feasible region will be in first quadrant.
Now, draw the graph of the line `3x+2y=20`

Put `(0,0)` in the inequation `3x+2yle20`
`3xx0+2xx0le20implies0le20` (True)
Therefore half plane contains the origin.

From equations `2x+y=12` and `3x+2y=20`
The point of intersection is `B(4,4)`
Therefore the feasible region is OABCO.
Its vertices are `O(0,0),A(6,0),B(4,4)` and `C(0,10)` we find the value of `Z` at these vertices.

The maximum value of `Z` is Rs. 32 at point `B(4,4),`. Therefore 4 pedestal lamp and 4 wooden shade should be made the manufacturer to obtain the maximum profit.
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