Question

A dietician wishes to mix 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 contains of per kg food is given below.

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

One kg of food $X$ costs rupees $24$ and one kg of food $Y$ costs rupees $36$. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.

Answer 1

Let $x$ be the number of units of food X and $y$ be the number of units of food Y be mixed to obtain the desired diet. Then LPP of the given problem is :
Minimise, $z=24x+36y$
Subject to the constraints,
$x+2y\ge 10,2x+2y\ge 12$ i.e., $x+y\ge 6$
$3x+y\ge 8$
$x\ge 0,y\ge 0$
We draw the lines $x+2y=10,x+y=6,3x+y=8$ and obtain the feasible region (unbounded and convex) shown in the figure.
Thus corner points are $A(0,8),B(1,5),C(2,4)$ and $D(10,0)$.
The values of $z$ (in rupees) at these points are given in the following table :

Corner Point	Objective Function $z=24x+36y$
$A(0,8)$	$z=24\times 0+36\times 8=288$
$B(1,5)$	$z=24\times 1+36\times 5=204$
$C(2,4)$	$z=24\times 2+36\times 4=192$
$D(10,0)$	$z=24\times 10+36\times 0=240$

As the feasible region is unbounded, we draw the graph of the half plane $6x+10y<52$ i.e., $3x+5y<26$ and note that there is no point common with the feasible region.
Therefore, $z$ has the minimum value of the minimum value is Rupees $192$.
It occurs at $C(2,4)$.
i.e., when $2kg$ of food $X$ and $4kg$ of food $Y$ are mixed to get the desired diet.