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 % of 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 Rs. 6 kg and $F_2$ costs Rs. 5 kg, determine how much of each at a minimum cost. What is the minimum cost ?

Answer 1

Let the farmer uses x kg of $F_1$ and y kg of $F_2$. We have construct the following table :

$\begin{array}{ccccc}Type& Quantity& Nitrogen& Phosphoric& Cost\\ & \left(inkg\right)& & acid& (inRs.)\\ F_1& x& \frac{10}{100}x=\frac{1}{10}x& \frac{6}{100}x& 6x\\ F_2& y& \frac{5}{100}y=\frac{1}{20}y& \frac{10}{100}y& 5y\\ Total& x+y& \frac{x}{10}+\frac{y}{20}& \frac{6x}{100}+\frac{10y}{100}& 6x+5y\\ Requirement\left(inkg\right)& & 14& 14& \end{array}$

Total cost of fertilizers, Z = 6x + 5y

So, our problem is to minimize Z = 6x + 5y .....(i)

Subject to constrains $\frac{x}{10}+\frac{y}{20}\ge 14\iff 2x+y\ge 280$ .......(ii)

$\frac{6x}{100}+\frac{10y}{100}\ge 14\iff 3x+5y\ge 700$ ........(iii)

$x\ge 0,y\ge 0$ .........(iv)

Firstly, draw the graph of the line 2x + y = 280

$\begin{array}{ccc}x& 0& 140\\ y& 280& 0\end{array}$

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

Secondly, draw the graph of the line

3x + 5y = 700

$\begin{array}{ccc}x& 0& \frac{700}{3}\\ y& 140& 0\end{array}$

Putting (0, 0) in the inequality 3x + 5y $\ge$ 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 $A(\frac{700}{3},0)$, B(100, 80)

and C(0, 280). The values of Z at these points are as follows :

$\begin{array}{cc}Cornerpoint& Z=6x+5y\\ A(\frac{700}{3},0)& 1400\\ B(100,80)& 1000\to Minimum\\ C(0,280)& 1400\end{array}$

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, 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.