Question

A new cereal was formed named Kellogg. It has a mixture of rice and bran that includes at least 38 grams of proteins and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs. 5 per kilogram and rice costs Rs. 4 per kilogram.

Answer 1

Hint: According to the question, the given problem is a Linear Programming Problem (LPP). Through this method, we can find out the minimum or maximum value in functions. We have to assign a variable for the minimum or maximum value which we are going to find out.

Formula used: \[z = 5x + 4y\]

Complete step by step answer:
First, we will assign variables to the amount of rice and bran. So, let us assume that we have bran of ‘x’ kg, and rice of ‘y’ kg.
We have to assign a variable to the minimum cost. So, let us assume that the minimum cost is ‘z’.
Now, according to the question, the value of ‘z’ will be determined by adding the product of total kilogram of bran with cost of per kilogram of bran and the product of total kilogram of rice with cost of per kilogram of rice. If we make an equation, we get:
\[z = 5x + 4y\]
Now, we can see that this is the Hypothesis of LPP.
Subject to constants are:
For protein:-
\[\left( {\dfrac{{80}}{{1000}}} \right)x + \left( {\dfrac{{100}}{{1000}}} \right)y \geqslant \left( {\dfrac{{88}}{{1000}}} \right)\]
When we simplify this equation, we get:
\[ \Rightarrow 20x + 25y \geqslant 22\]
This will be now denoted as our equation (1).
For iron:-
\[\left( {\dfrac{{40}}{{1000}}} \right)x + \left( {\dfrac{{30}}{{1000}}} \right)y \geqslant \left( {\dfrac{{36}}{{1000}}} \right)\]
When we simplify this, we get:
\[ \Rightarrow 20x + 15y \geqslant 18\]
This will be denoted as our equation (2)
Our next equation will be \[x \geqslant 0\,and\,y \geqslant 0\].
This is our equation (3).
Now, with the help of these three equations, we will make a graph.
In equation (1),
\[ \Rightarrow 20x + 25y \geqslant 22\]
When \[x = 0\], then \[y = \dfrac{{22}}{{25}}\].
When \[y = 0\], then \[x = \dfrac{{22}}{{20}} - 1.1\]
In equation (2),
\[ \Rightarrow 20x + 15y \geqslant 18\]
When \[x = 0\], then \[y = \dfrac{{18}}{{15}} = 1.2\]
When \[y = 0\], then \[x = \dfrac{{18}}{{20}}\]
When we plot a graph, we get:

Our main part is finding out the intersection point. When we solve both the equations (1) and (2), then we get the intersection point. When we solve, we get:
\[
  \;\;\;20x + 25y = 22 \\
   - \;\underline {20x + 15y = 18} \\
  \;\;\;\;\;\;\;\;\;\;\;10y = 4 \\
   \Rightarrow y = \dfrac{4}{{10}} = 0.4\;{\text{and}}\;x = 0.6 \\
 \]
So, the coordinates for intersection points are \[(0.6,\,0.4)\].
Now, we will make a table for the corner points. We had to minimize the equation \[z = 5x + 4y\].
Now, to make the table, we will consider only the maximum points from the graph, and we get:
\[A(1.1,\,0)\,and\,B(0.6,\,0.4)\,and\,C(0,\,1.2)\]
For point ‘A’:
\[z = 5 \times 1.1 + 4 \times 0\]
\[ \Rightarrow z = 5.5\]
For point ‘B’:
\[z = 5 \times 0.6 + 4 \times 0.4\]
\[ \Rightarrow z = 4.6\]
For point ‘C’:
\[z = 5 \times 0 + 4 \times 1.2\]
\[ \Rightarrow z = 4.6\]
So, the minimum value for ‘z’ is \[4.6\].
We know that we had to find the minimum cost. Our minimum cost was denoted as ‘z’. Therefore, the minimum cost is Rs. \[4.6\] per kg.


Note: Generally, the question or problem is given in the form of a function. The use of LPP is normally for management purposes or decision-making problems. We use the graph when only two variables are given. If more than two variables are given, then the graph making is not suitable. So, we have to use the Simplex Method.