Question

There are two types of fertilizers $F_1$ and $F_2$. $F_1$ consists of $10$% nitrogen and $6$% phosphoric acid and $F_2$ consists of $5$% nitrogen and $10$% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least $14$kg of nitrogen and $14$kg of phosphoric acid for her crop. If $F_1$ cost Rs.$6$/kg and $F_2$ costs Rs.$5$/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer 1

Let the farmer buy $x$kg of fertilizer $F_1$ and $y$ kg of fertilizer $F_2$. Therefore, $x\ge 0$ and $y\ge 0$
The given information can be compiled in a table as follows.

	Nitrogen (%)	Phosphoric Acid (%)	Cost (rs/kg)
$F_1\left(x\right)$	$10$	$6$	$6$
$F_2\left(y\right)$	$5$	$10$	$5$
Requirement	$14$	$14$

$F_1$ consists of $10$% nitrogen and $F_2$ consists of $5$% nitrogen. However, the farmer requires at least $14$kg of nitrogen.
$\therefore$ $10$% of $x$ $+$ $5$% of $y\ge 14$
$\frac{x}{10}+\frac{y}{20}\ge 14$
$2x+y\ge 280$
$F_1$ consists of $6$% phosphoric acid and $F_2$ consists of $10$% phosphoric acid. However, the farmer requires at least $14$kg of phosphoric acid
$\therefore$ $6$% of $x$ $+$ $10$% of $y\ge 14$
$\frac{6x}{100}+\frac{10y}{100}\ge 14$
$3x+56y\ge 700$
Total cost of fettilizers, $Z=6x+5y$
The mathematical formulation of the given problem is
Minimise $Z=6x+5y.....\left(1\right)$
subject to the constraints
$2x+y\ge \mathrm{280......}\left(2\right)$
$3x+5y\ge \mathrm{700.....}\left(3\right)$
$x,y\ge \mathrm{0......}\left(4\right)$
The feasible region determined by the system of constraints is as shown
It can be seen that the feasible region is unbounded.
The corner points are $A\left(\frac{700}{3}\right),B(100,80)$ and $C(0,280)$
The values of $Z$ at these points are as follows

Corner point	$Z=6x+5y$
$A(\frac{700}{3},0)$	$1400$
$B(100,80)$	$1000$	$\to$ Minimum
$C(0,280)$	$1400$

As the feasible region is unbounded, therefore, $1000$ may or may ot be the minimum value of$Z$
For this, we draw a graph of the inequality $6x+5y<1000$ and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with
$6x+5y<1000$
Therefore, $100$kg of fertilizer $F_1$ and $80$ kg of fertilizer $F_2$ should be used to minimize the cost. The minimum cost is Rs.$1000$.