Question

A dietician wishes to mix together two kinds of food $X$ and $Y$ in such a way that the mixture contains at least $10$ units of vitamin $A$, $12$ units of vitamin $B$ and $8$ units of vitamin $C$. The vitamin content of one kg of food is given below:

Food	Vitamin A	Vitamin B	Vitamin C
$X$	$1$	$2$	$3$
$Y$	$2$	$2$	$1$

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

Answer 1

Let the mixture contain $x$ kg of food $X$ and $y$ kg of food $Y$
The mathematical formulation of the given problem is as follows.
Minimize $z=16x+20y......\left(1\right)$

subject to the constraints,
$x+2y\ge 10......\left(2\right)$
$x+y\ge 6.........\left(3\right)$
$3x+y\ge 8.......\left(4\right)$
$x,y\ge \mathrm{0..........}\left(5\right)$

The feasible region determined by the system of constraints is as follows:
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 corner points are as follows.

Corner point	$z=16x+20y$
$A(10,0)$	$160$
$B(2,4)$	$112$	$\to$ Minimum
$C(1,5)$	$116$
$D(0,8)$	$160$

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, $16x+20y<112$ or $4x+5y<28$ 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 $4x+5y<28$
Therefore, the minimum value of $z$ is $112$ at $(2,4)$.

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