Question

Reshma wishes to mix two types of food $P$ and $Q$ in such a way that the vitamin contents of the mixture contain at least $8units$ of vitamin $A$ and $11units$ of vitamin $B$. Food $P$ costs $Rs.60/kg$ and Food $Q$ costs $Rs.80/kg$. Food $P$ contains $3units/kg$ of Vitamin $A$ and $5units/kg$ of Vitamin $B$ while food $Q$ contains $4units/kg$ of Vitamin $A$ and $2units/kg$ of vitamin $B$. Determine the minimum cost of the mixture.

Answer 1

Let's assume that mixture contains $Xkg$ of food P type and $Ykg$ of food Q type.

Cost of 1 kg of food P $=60Rs$

Cost of 1 kg of food Q $=80Rs$

So, Total cost (C) $=60X+80YRs...\left(1\right)$

Now, food P contains $3units/kg$ of vitamin A and food Q contains $4units/kg$ of vitamin A.

So, Total Vitamin A contains $=3X+4Y$

Since minimum requirement of Vitamin A is 8 units.

$\therefore 3X+4Y\ge 8...\left(2\right)$

Also, food P contains $5units/kg$ of vitamin B and food Q contains $2units/kg$ of vitamin B.

So, Total Vitamin B contains $=5X+2Y$

Since the minimum requirement of Vitamin B is 11 units.

$\therefore 5X+2Y\ge 11...\left(3\right)$

Since the amount of food can never be negative.

So, $X\ge 0,Y\ge 0...\left(4\right)$

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

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

Corner points	Value of $Z=60X+80Y$
$(0,\frac{11}{2})$	$440$ (Maximum)
$(2,\frac{1}{2})$	$160$ (Minimum)
$(\frac{8}{3},0)$	$160$ (Minimum)

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 $Z<160$ to check if there exist some points in the feasible region where value can be less than 160.

$\Rightarrow 60X+80Y<160$

$\Rightarrow 3X+4Y<8$.

Since there is no common points between the feasible region and the region which contains $Z<160$. So 160 is the minimum cost.