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 14kg of nitrogen and 14kg of phosphoric acid for her crop. If F1 cost Rs.6/kg and F2 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

Hint: To solve this question, we should know the graphical method to solve optimisation problems. Let us assume the number of kgs of ${{F}_{1}}$ and ${{F}_{2}}$ be $x$ and $y$ respectively. We can write the total amount of nitrogen of phosphoric acid from the given percentages. For example the amount of nitrogen supplied is $\dfrac{10}{100}x+\dfrac{5}{100}y$. Likewise, we can write for phosphoric acid. Using the minimum values given, we can plot the area which suits the requirements. We need to minimise the cost which is $6x+5y$. Using the graph, we can get the minimum value of the cost and the other values asked in the question.

Complete step by step answer:

Let us assume the number of kgs of ${{F}_{1}}$ and ${{F}_{2}}$ be $x$ and $y$ respectively. We are given that ${{F}_{1}}$ consists of 10% nitrogen and 6% phosphoric acid and ${{F}_{2}}$ consists of 5% nitrogen and 10% phosphoric acid. Using this data, we can get the amount of nitrogen and phosphoric acid supplied to the field when ${{F}_{1}}$ and ${{F}_{2}}$ be $x$ and $y$kgs respectively.
Nitrogen $=\dfrac{10}{100}x+\dfrac{5}{100}y=\dfrac{x}{10}+\dfrac{y}{20}$
Phosphoric acid $=\dfrac{6}{100}x+\dfrac{10}{100}y=\dfrac{3x}{50}+\dfrac{y}{10}$
We are given the minimum requirement of nitrogen and phosphoric acid as 14 kgs for the field. Using this, we can write that
$\begin{align}
  & \dfrac{x}{10}+\dfrac{y}{20}\ge 14 \\
 & \Rightarrow \dfrac{2x+y}{20}\ge 14 \\
 & \Rightarrow 2x+y\ge 280\to \left( 1 \right) \\
\end{align}$
$\begin{align}
  & \dfrac{3x}{50}+\dfrac{y}{10}\ge 14 \\
 & \Rightarrow \dfrac{3x+5y}{50}\ge 14 \\
 & \Rightarrow 3x+5y\ge 700\to \left( 2 \right) \\
\end{align}$
We can infer that both the inequalities are unbounded. Let us consider
$\begin{align}
  & 2x+y=280\to \left( 3 \right) \\
 & 3x+5y=700\to \left( 4 \right) \\
\end{align}$
And solve them to draw the graph. Multiplying the equation-3 by 5 and subtracting equation-4 from equation-3, we get
$5\left( 2x+y=280 \right)$
$\begin{align}
  & \underline{\begin{align}
  & 10x+5y=1400 \\
 & -\left( 3x+5y=700 \right) \\
\end{align}} \\
 & 7x=700 \\
 & \Rightarrow x=100 \\
\end{align}$
Using this in equation-3, we get
$\begin{align}
  & 2\left( 100 \right)+y=280 \\
 & \Rightarrow y=80 \\
\end{align}$
When we plot both the inequalities 1 and 2, we get an unbounded region. Plotting them on a graph, we get

The shaded region is the region which suits the requirement given.
The cost function is given by $6x+5y$. We need to minimise the cost which is $6x+5y$.
The cost when $x=100,y=80$ which is the point of intersection is given by
Cost = $6\times 100+5\times 80=1000$.
Now, let us consider the region where the cost is less than 1000. It is
$6x+5y\le 1000$.
Plotting it on the graph, we get

The green region is the region which has the value of the cost less than 1000. We can see from the graph that there is no other common point or region between the green region and the minimum requirement region. So, the point $x=100,y=80$, gives the minimum cost satisfying the requirements. The minimum cost is Rs. 1000. The values of the nutrients supplied are
Nitrogen $=\dfrac{100}{10}+\dfrac{80}{20}=14kg$
Phosphoric acid $=\dfrac{3\times 100}{50}+\dfrac{80}{10}=14kg$
$\therefore $The minimum cost for the given requirements is Rs. 1000. The nutrients supplied at this cost are 14 kgs each of nitrogen and phosphoric acid.

Note:
If we observe the result, we can observe that we got the minimum cost when both the constraint lines intersected. This is a general result in these types of questions. The other points which we should check in these types of questions are the points where the lines meet the x and y axis. As the values of x, y are positive, these points can be crucial while we are doing an optimisation problem.