Let `x` bags of brand `P` and `y` of brand `Q` are used by fruit grower.
Minimise `Z=3x+3.5y`……………1
and constraints `x+2yge240`………………2
`3x+1.5yge270`………………….3
`1.5x+27le310`……………..4
`xge0,yge0`.......................5
First draw the graph of the line `x+2y=240`
Put `(0,0)` in the inequation `x+2yge240`
`0+2xx0ge240implies0ge240` (False)
Thus, the half plane does not contain origin.
Now, draw the graph of the line `3x+1.5y=270`
Put `(0,0)` in the inequations `3x+1.5yge270`
`3xx0+1.5xx0ge270implies0ge270` (False)
Thus, the half plane does not contain origin.
Now, draw the graph of the line `1.5x+2y=310`
PUt `(0,0)` in the inequation `1.5x+2yle310`
`1.5xx0+2xx0le310implies0le310` (True)
Thus, the half plane does not contain origin.
Since `x,yge0`
So the feasible region will be in first quadrant.
The point of intersection of equations `1.5x+2y=310` and `x+2y=240` is `A(140,50)`. Similarly the point of intersection of the equation `3x+1.5y=270` and `1.5x+2y=310` is `B(20,140)`.
`:.` Feasible region is ABCA.
It vertices are `A(140,50),B(20,140)` and `C(40,100)` at which we find the value of `Z`.
Therefore, the minimum value of `Z` is 470 at point `C(40,100)`. Minimum nitrogen to be mixed in garden is 470 kg for which 40 bags of brand `P` and 100 bats of brand `Q` should be used.