Question

A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of 1 kg food is given below.

$\begin{array}{cccc}Food& VitaminA& VitaminB& VitaminC\\ X& 1& 2& 3\\ Y& 2& 2& 1\end{array}$

1 kg of food X costs of Rs. 16 and 1 kg of food Y costs Rs. 20. Find the least cost of the mixture which will produce the required diet ?

Answer 1

Let the dietician mixes x kg of food X and y kg of food Y. We construct the following table:

$\begin{array}{cccccc}Food& Amount& VitaminA& VitaminB& VitaminC& Cost\\ X& xkg& x& 2x& 3x& 16x\\ Y& ykg& 2y& 2y& y& 20y\\ Total& (x+y)kg& x+2y& 2x+2y& 3x+y& 16x+20y\\ Minimum& & 10& 12& 8& \\ Requires& & & & & \end{array}$

So, our problem is to minimize Z = 16x + 20y .........(i)

Subject to constraints x + 2y $\ge$ 10 ...........(ii)

$2x+2y\ge 12\iff x+y\ge 6$ ........(iii)

3x + y $\ge$ 8 .......(iv)

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

Firstly draw the graph of the line x + 2y = 10

$\begin{array}{ccc}x& 0& 10\\ y& 5& 0\end{array}$

Putting (0, 0) in the inequality x + 2y $\ge$ 10, we have

$0+2\times 0\ge 10\Rightarrow 0\ge 10$ (which is false)

So, the half plane is away from the origin.

Secondly draw the graph of the line x + y = 6

$\begin{array}{ccc}x& 0& 6\\ y& 6& 0\end{array}$

Putting (0, 0) in the inequality x + y $\ge$ 6 we have

$0+0\ge 6\Rightarrow 0\ge 6$ (which is false)

So, the half plane is a way from the origin.

Thirdly draw the graph of the line 3x + y = 8

$\begin{array}{ccc}x& 0& \frac{8}{3}\\ y& 8& 0\end{array}$

Putting (0, 0) in the inequality 3x + y $\ge$ 8, we have

$3\times 0+0\ge 8\Rightarrow 0\ge 8$ (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.

On solving equations x + y = 6 and x + 2y = 10, we get B(2, 4)

Similarly, solving the equations 3x + y = 8 and x + y = 6, we get C(1, 5).

The corner points of the feasible region are A(10, 0), B(2, 4), C(1, 5) and D(0, 8). The values of Z at these points are as follows :

$\begin{array}{cc}Cornerpoint& Z=16x+20y\\ A(10,0)& 160\\ B(2,4)& 112\to Minimum\\ C(1,5)& 116\\ D(0,8)& 160\end{array}$

As the feasible region is unbounded, therefore 112 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, 16x + 20 y < 112 or 4x + 5y < 28 and check, whether the resulting half plane has points common with the feasible region or not.

It can be seen that the feasible region has no common point with 4x + 5y < 28.

Therefore, the minimum value of Z is 112 at B(2, 4).

Thus, the mixture should contain 2 kg of food X and 4 kg of food Y, The minimum cost of the mixture is Rs. 112.