Question

A farmer mixes two brands P and Q of cattle feed. Brand P costing Rs.250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q, costing Rs.200 per bag contain 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirement of nutrients A, B and C are 18 units, 45 units and 24 units respectively.
Determine the number of bags of each brand which should be mixed in order to produce a mixture having minimum cost per bag? What is the minimum cost of the mixture per bag?

Answer 1

Hint: This is a question of linear programming problem. For solving this, we will first tabulate the data. Using the given information, we will identify our objective function z and from the data, we will identify the constraints. Then we will draw graphs for the constraints (inequalities) and find their common region. From the common region, we will find the corner points. Then we will put these points one by one in the objective function z. Minimum value of z will give us minimum cost and the point (value of x,y) where cost is minimum.

Complete step by step answer:
Let us first suppose that the mixture contains x bags of brand P and y bags of brand Q.
Brand P contains 3 units of element A, 2.5 units of element B and 2 units of element C.
Brand Q contains 1.5 units of element A, 11.25 units of element B and 3 units of element C.
Minimum requirement of element A is 18 units, element B is 45 units and element C is 24 units. Tabulating this data we have,

Brand	Number of boys	Element A	Element B	Element C
P	x	3 units	2.5 units	2 units
Q	y	1.5 units	11.25 units	3 units

Least requirement

18 units

45 units

24 units

Now let us form the constraints using the table,
For element A we have,
$3x+1.5y\ge 18$ (Mixture should have atleast 18 units of element A).
Dividing both sides by 1.5 we have ${\displaystyle \frac{3}{1.5}}x+{\displaystyle \frac{1.5y}{1.5y}}\ge {\displaystyle \frac{18}{1.5}}$.
Simplifying we get $2x+y\ge 12\cdots \cdots \cdots \left(1\right)$.
For element B, we have $2.5x+11.25y\ge 45$.
Dividing by 1.25 we get ${\displaystyle \frac{2.5}{1.25}}x+{\displaystyle \frac{11.25}{1.25}}y\ge {\displaystyle \frac{45}{1.25}}$.
$$\left[\begin{array}{rl}& {\displaystyle \frac{2.5}{1.25}}={\displaystyle \frac{25}{10}}\times {\displaystyle \frac{100}{125}}=2\\ & {\displaystyle \frac{11.5}{1.25}}={\displaystyle \frac{1125}{125}}\times {\displaystyle \frac{100}{100}}=9\\ & {\displaystyle \frac{45}{1.25}}={\displaystyle \frac{45}{125}}\times 100=36\end{array}\right]$$
Simplifying we get $2x+9y\ge 36\cdots \cdots \cdots \left(2\right)$.
For element C we have $2x+3y\ge 24\cdots \cdots \cdots \left(3\right)$.
Also we know that the number of bags cannot be negative so $x\ge 0,y\ge 0\cdots \cdots \cdots \left(4\right)$.
Now we need to minimize the cost. Let us suppose objective function is z which we need to minimize.
Cost of brand P is Rs.250 and cost of brand Q is Rs.200.
Hence we need to minimize the function.
$z=250x+200y$.
Now objective function z is subject to the constraints given by (1), (2), (3), (4). So our LPP looks like this,
Minimize $z=250x+200y$ subject to the constraints,
$$\begin{array}{rl}& 2x+y\ge 12\\ & 2x+9y\ge 36\\ & 2x+3y\ge 24\\ & x\ge 0,y\ge 0\end{array}$$
Now let us draw a graph for all the constraints and find the common point.
Coordinates for inequalities can be formed as,
For $2x+y\ge 12$ taking $2x+y=12$.
$\begin{array}{rl}& x=0\Rightarrow y=12\\ & y=0\Rightarrow x=6\end{array}$

x	6	0
y	0	12

For $2x+9y\ge 36$ taking $2x+9y=36$.
$\begin{array}{rl}& x=0\Rightarrow y=4\\ & y=0\Rightarrow x=18\end{array}$

x	0	18
y	4	0

For $2x+3y\ge 24$ taking $2x+3y=24$.
$\begin{array}{rl}& x=0\Rightarrow y=8\\ & y=0\Rightarrow x=12\end{array}$

x	0	12
y	8	0

Our graph looks like this,

From the graph we can see that the corner points are A(0,12), B(3,6), C(9,2) and D(18,0).
Let us use this value in z, and find minimum value of z
For A(0,12) $z=250\left(0\right)+200\left(12\right)=2400$.
For B(3,6) $z=250\left(3\right)+200\left(6\right)=750+1200=1950$.
For C(9,2) $z=250\left(9\right)+200\left(2\right)=2250+400=2650$.
For D(18,0) $z=250\left(18\right)+200\left(0\right)=4500$.
As we can see, we have minimum z at B(3,6).
Therefore, the minimum cost is Rs.1950.
Required values of x and y are 3, 6 respectively.

Hence, the cost will be minimum if the mixture contains 3 bags of brand P and 6 bags of brand Q.
Also the minimum cost will be Rs.1950.

Note: Students should note that the region is unbounded. Maximum value of such LPP cannot be found. To check if the obtained value is minimum they can draw the inequality 250x+200y<1950 on graph and check for any common point between feasible region and the inequality. If there does not exist any point the answer is correct.