Question

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

Answer 1

Let the mixture contain $xkg$ of food $P\text{and}ykg$ of food $Q$
Hence, combining the constraints:
Minimise $Z=60x+80y$
Subject to constraints:
$3x+4y\ge 8$
$5x+2y\ge 11$
$x\ge 0,y\ge 0$

$3x+4y=8$
$\begin{array}{ccc}x& 0& \frac{8}{3}\\ y& 2& 0\end{array}$

$5x+2y=11$
$\begin{array}{ccc}x& 0& \frac{11}{5}\\ y& \frac{11}{5}& 0\end{array}$
$\begin{array}{cc}\text{Corner Points}& \text{Value of}Z=60x+80y\\ (0,5.5)& 440\\ (2,\frac{1}{2})& 160\\ (\frac{8}{3},0)& 160\end{array}$
Since the feasible region is unbounded.
Hence, 160 may or may not be minimum value of$Z$.

So, we need to graph inequality:

$60x+80y<160$
$\Rightarrow 3x+4y<8$
$5x+2y=11$
As, there is no common points between the feasible region and the inequality.
$\therefore 160$ is the minimum value of $Z$.
Since, $Z$ is minimum at two points
$(\frac{8}{3},0)\&(2,\frac{1}{2})$

Therefore, minimum cost of mixture is $Rs.160$ at all points joining $(\frac{8}{3},0)\&(2,\frac{1}{2})$