Question

There are two types of fertilisers $F_1$ 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 atleast $14kg$ of nitrogen and $14kg$ of phosphoric acid for her crop. If $F_1$ costs ₹ $6/kg$ and $F_2$ costs $₹5/kg$, determine how much of each type of fertiliser should be used, so that nutrient requirements are met at a minimum cost?

• A. $100kg$ of fertiliser $F_1$ and $80kg$ of fertiliser $F_2$
• B. $80kg$ of fertiliser $F_1$ and $100kg$ of fertiliser $F_2$
• C. $100kg$ of each fertilisers $F_1$ and $F_2$
• D. $80kg$ of each fertilisers $F_1$ and $F_2$

Answer 1

Answer: (A)

Let the farmer uses $xkg$ of $F_1$ and $ykg$ of $F_2$. We have construct the following table

Type	Quantity(in $kg$ )	Nitrogen	Phosphoric acid	Cost(in ₹)
$F_1$	x	$\frac{10}{100}x=\frac{1}{10}x$	$\frac{6}{100}x$	6x
$F_2$	y	$\frac{5}{100}y=\frac{1}{20}y$	$\frac{10}{100}y$	5y
Total	x+y	$\frac{x}{10}+\frac{y}{20}$	$\frac{6x}{100}+\frac{10y}{100}$	$6x+5y$
Require ment(in $kg$ )		14	14

Total cost of fertilizers, $Z=6x+5y$
So, our problem is to minimise $Z=6x+5y$...(i)
Subject to constraints are
$\frac{x}{10}+\frac{y}{20}\ge 14\iff 2x+y\ge 280$...(ii)
$\frac{6x}{100}+\frac{10y}{100}=14\iff 3x+5y\ge 700$...(iii)
$x\ge 0,y\ge 0$...(iv)
Firstly, draw the graph of the line $2x+y=280$

x	0	140
y	280	0

Putting $(0,0)$ in the inequality $2x+y\ge 280$, we have
$2\times 0,0\ge 280$
$\Rightarrow 0\ge 280$(which is false)
So, the half plane is away from the origin.
Since, $x,y\ge 0$
So, the feasible region lies in the first quadrant.
Secondary, draw the graph of the line
$3x+5y=700$

x	0	700/3
y	140	0

Putting $(0,0)$ in the inequality $3x+5y\ge 700$, we have
$3\times 0+5\times 0\ge 700$
$\Rightarrow 0\ge 700$(which is false)

So, the half plane is away from the origin.
On solving the equations $2x+y=280$ and $3x+5y=700$, we get $B(100,80)$.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are $A\left(\frac{700}{3},0\right)$, $B(100,80)$ and $C(0,280)$. The values of $Z$ at these points are as follows

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

As the feasible region is unbounded, therefore 1000 may or may not 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, $100kg$ of fertilizer $F_1$ and $80kg$ of fertilizer $F_2$ should be used to minimise the cost. The minimum cost is $₹1000$.