Question

A fruit grower can use two types of fertilizer in his garden, brand $P$ and brand $Q$.The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least $240kg$ of phosphoric acid, at least $270kg$ of potash and at most $310kg$ 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?

Kg per bag
	Brand $P$	Brand $Q$
Nitrogen	$3$	$3.5$
Phosphoric acid	$1$	$2$
Potash	$3$	$1.5$
Chlorine	$1.5$	$2$

Answer 1

Let's assume that bags of brand P needed be $X$ and brand Q be $Y$

Since bag of brand P contains 1 kg and bag of brand Q contains 2 kg of Phosphoric acid. Also, garden needs at least 240 kg of phosphoric acid.

$\therefore X+2Y\ge 240...\left(1\right)$

Since bag of brand P contains 3 kg and bag of brand Q contains 1.5 kg of potash. Also, garden needs at least 270 kg of potash.

$\therefore 3X+1.5Y\ge 270$

$\Rightarrow 2X+Y\ge 180...\left(2\right)$

Since bag of brand P contains 1.5 kg and bag of brand Q contains 2 kg of chlorine. Also, garden needs at most 310 kg of chlorine.

$\therefore 1.5X+2Y\le 310...\left(3\right)$

Since count of bags can't be negative.

$\therefore X\ge 0,Y\ge 0...\left(4\right)$

Now, Bag of brand P contains 3 kg of Nitrogen and Bag of brand Q contains 3.5 kg of Nitrogen.

So, total Nitrogen contents $Z=3X+3.5Y$

We have to minimise the amount of Nitrogen added to the garden.

After plotting all the constraints given by equation (1), (2), (3) and (4) we get the feasible region as shown in the image.

Corner points	Value of $Z=3X+3.5Y$
A (20, 140)	550
B (40, 100)	470 (Minimum)
C (140, 50)	595

Hence, to minimise the amount of nitrogen added to the garden, grower should use $40$ bags of brand P and $100$ bags of brand Q. In this case minimum amount of nitrogen added to the garden will be $470Kg$