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.