Question

There are two types of fertilisers $F_1$ and $F_2$. $F_1$ consists of $10\%$ nitrogen and $6\%$ phosphoric acid, $F_2$ consists of $5\%$ nitrogen and $10\%$ phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least $14\text{kg}$ of nitrogen and $14\text{kg}$ of phosphoric acid for her crop. If $F_1$ costs Rs $6/\text{kg}$ and $F_2$ costs Rs $5/\text{kg}$, then the minimum cost of mixture which will produce the required nutrients is

Answer 1

Suppose $x\text{kg}$ of fertiliser $F_1$ and $y\text{kg}$ of fertiliser $F_2$ is used to reach the minimum requirement of the soil. Then
Minimum cost (in Rs) $=6x+5y$
Let $Z=6x+5y$

We now have the following mathematical model for the given problem.
Minimize $Z=6x+5y\cdots \left(i\right)$
subject to the constraints :
$2x+y\ge 280\cdots \left(ii\right)$ (nitrogen requirement constraint)
$3x+5y\ge 700\cdots \left(iii\right)$ (phosphoric requirement constraint)
$x\ge 0,y\ge 0\cdots \left(iv\right)$ (non-negative constraint)

The feasible region (shaded) determined by the linear inequalities $\left(i\right)$ to $\left(iv\right)$ is shown in the figure. Note that the feasible region is unbounded.


Let us evaluate the objective function $Z$ at each corner point as shown below :

Corner point	$Z=6x+5y$
$A(0,280)$	$1400$
$B(100,80)$	$1000\to \text{Minimum}$
$C(233.33,0)$	$1400$

We find that minimum value of $Z$ is Rs $1000$ at $(100,80).$
Hence, farmer should buy $100\text{kg}$ of fertiliser $F_1$ and $80\text{kg}$ of fertiliser $F_2$ to meet the requirements at minimum cost and the minimum cost will be Rs $1000$.