Q. 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 $14 kg$ of nitrogen and $14 kg$ 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?
Linear Programming
Solution:
Let the farmer uses $x kg$ of $F_1$ and $y kg$ 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
$\frac{x}{10}+\frac{y}{20}$
$\frac{6 x}{100}+\frac{10 y}{100}$
$6 x+5 y$
Require ment(in $kg$ )
14
14
Total cost of fertilizers, $Z=6 x+5 y$
So, our problem is to minimise $Z=6 x+5 y$...(i)
Subject to constraints are
$\frac{x}{10}+\frac{y}{20} \geq 14 \Leftrightarrow 2 x+y \geq 280 $...(ii)
$\frac{6 x}{100}+\frac{10 y}{100}=14 \Leftrightarrow 3 x+5 y \geq 700 $...(iii)
$x \geq 0, y \geq 0$...(iv)
Firstly, draw the graph of the line $2 x+y=280$
x
0
140
y
280
0
Putting $(0,0)$ in the inequality $2 x+y \geq 280$, we have
$2 \times 0,0 \geq 280 $
$\Rightarrow 0 \geq 280$(which is false)
So, the half plane is away from the origin.
Since, $ x, y \geq 0$
So, the feasible region lies in the first quadrant.
Secondary, draw the graph of the line
$3 x+5 y=700$
x
0
700/3
y
140
0
Putting $(0,0)$ in the inequality $3 x+5 y \geq 700$, we have
$ 3 \times 0+5 \times 0 \geq 700$
$\Rightarrow 0 \geq 700$(which is false)
So, the half plane is away from the origin.
On solving the equations $2 x+y=280$ and $3 x+5 y=700$, 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
$B(100,80)$
$1000 \rightarrow$ 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, $6 x+5 y<1000$ 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 $6 x+5 y<1000$.
Therefore, $100 kg$ of fertilizer $F_1$ and $80 kg$ of fertilizer $F_2$ should be used to minimise the cost. The minimum cost is $₹ 1000$.
| 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 | $\frac{x}{10}+\frac{y}{20}$ | $\frac{6 x}{100}+\frac{10 y}{100}$ | $6 x+5 y$ |
| Require ment(in $kg$ ) | 14 | 14 |
| x | 0 | 140 |
| y | 280 | 0 |
| x | 0 | 700/3 |
| y | 140 | 0 |
| Corner point | $Z=6x+5y$ |
|---|---|
| $A\left(\frac{700}{3}, 0\right)$ | 1400 |
| $B(100,80)$ | $1000 \rightarrow$ Maximum |
| $C(0,280)$ | 1400 |