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 contents of one kg food is given below.

Food	$\text{Vitamin}A$	$\text{Vitamin}B$	$\text{Vitamin}C$
$X$	$1$	$2$	$3$
$Y$	$2$	$2$	$1$

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

Answer 1

Let's assume that mixture contains $Xkg$ of food $X$ and $Y\text{kg}$ of food $Y$.
Cost of $1\text{kg}$ of food $X=\text{Rs}16$
Cost of $1\text{kg}$ of food $Y=\text{Rs}20$
So, Total cost
$\left(Z\right)=16X+20Y\cdots \left(i\right)$
Now, food $X$ contains $1$ units and food $Y$ contains $2$ units of Vitamin $A$.
So, Total Vitamin
$A=X+2Y$ units
Since, minimum requirement of Vitamin $A$ is $10$ units.
$\therefore X+2Y\ge 10\cdots \left(ii\right)$
Again, food $X$ contains $2$ units and food $Y$ contains $2$ units of Vitamin $B$.
So, Total Vitamin
$B=2X+2Y$ units
Since, minimum requirement of Vitamin $B$ is $12$ units.
$\therefore 2X+2Y\ge 12$
$\Rightarrow X+Y\ge 6$
Also, food $X$ contains $3$ units and food $Y$ contains $1$ units of Vitamin $C$.
So, Total Vitamin
$C=3X+Y$ units
Since, minimum requirement of Vitamin $C$ is $8$ units.
$\therefore 3X+Y\ge 8\cdots \left(iv\right)$
Since, amount of food can never be negative.
So, $X\ge 0,Y\ge 0\cdots \left(v\right)$
We have to minimise the cost of mixture given in equation $\left(i\right)$ subject to the constraints given in $\left(ii\right),\left(iii\right),\left(iv\right)$ and $\left(v\right)$. After plotting all the constraints, we get the feasible region as shown in the figure.
Corner points Value of $Z=16X+20Y$

Corner points	$Z=16X+20Y$
$A(0,8)$	160
$B(1,5)$	116
$C(2,4)$	112 (minimum)
$D(10,0)$	160

Now, since region is unbounded.
Hence we need to confirm that minimum value obtained through corner points is true or not.
Now, plot the region $Z<112$ to check if there exist some points in feasible region where value can be less than $112$.
$\Rightarrow 16X+20Y<112$
$\Rightarrow 4X+5Y<28$
Since there is no common points between the feasible region and the region which contains $Z<112$.
So $Rs112$ is the minimum cost.