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	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's assume that mixture contains $Xkg$ of food X and $Ykg$ of food Y.

Cost of 1 kg of food X $=16Rs$

Cost of 1 kg of food Y $=20Rs$

So, Total cost (C) $=16X+20YRs...\left(1\right)$

Now, food X contains $1units$ and food Y contains $2units$ of Vitamin A.

So, Total Vitamin A $=X+2Yunits$

Since, minimum requirement of Vitamin A is 10 units.

$\therefore X+2Y\ge 10...\left(2\right)$

Again, food X contains $2units$ and food Y contains $2units$ of Vitamin B.

So, Total Vitamin B $=2X+2Yunits$

Since, minimum requirement of Vitamin B is 12 units.

$\therefore 2X+2Y\ge 12$

$\Rightarrow X+Y\ge 6...\left(3\right)$

Also, food X contains $3units$ and food Y contains $1units$ of Vitamin C.

So, Total Vitamin C $=3X+Yunits$

Since, minimum requirement of Vitamin C is 8 units.

$\therefore 3X+Y\ge 8...\left(4\right)$

Since, amount of food can never be negative.

So, $X\ge 0,Y\ge 0...\left(5\right)$

We have to minimise the cost of mixture given in equation (1) subject to the constraints given in (2),(3), (4) and (5).

After plotting all the constraints, we get the feasible region as shown in the image.

Corner points	Value of $Z=16X+20Y$
A (0,12)	160
B (3,6)	116
C (9,2)	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 $Z<112$ (See the image). So $112Rs$ is the minimum cost.