A fruit grower canuse two types of fertilizer in his garden brand `P` and brand `Q`. The amounts (in kg) of nitrogen phosphoric acid, potash, and chlorine a bag of each brand are given in the stable. Tests indicate that the garden needs at least 240kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

Explanation

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.