Let the manufacturer produces `x` packets of nuts and `y` packets of bolts, then
`:.` Maximise `Z=17.5x+7y`……………….1
and constraints `x+3yle12`………………………2
`3x+yle12`………………….3
`xge0,yge0`………………...4
First draw the graph of the line `x+3y=12`.
Put `(0,0)` in the inequation `x+3yge12` ,
`0+3xx0le10implies0le12` (True)
Therefore half plane contains the origin. Since `x,yge0`. So the feasible region is in first quadrant,
Now draw the graph of the line `3x+y=12`
Put `(0,0)` in the inequation `3x+yle12`,
`3xx0+0le12gt0le12` (True)
Therefore, the half plane contains the origin.
From the equations `x+3y=12` and `3x+y=12`, the point of intersection is `B(3,3)`.
`:.` The feasible region is OABCO whose vertices are `O(0,0),A(4,0),B(3,3)` and `C(0,4)`. Now, we find the value of `Z` at these points,
The value of `Z` is maximum Rs. 73.50 at point `(3,3)`. Therefore, the maximum profit of Rs. 73.50 can be obtained when 3 packets of nuts and 3 packets of bolts are produced per day.