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 $8$ $units$ of vitamin $A$ and $11$ $units$ of vitamin $B$. Food $P$ costs Rs $60/kg$ and Food $Q$ costs Rs $80/kg$. Food $P$ contains $3$ $units/kg$ of Vitamin $A$ and $5$ $units/kg$ of Vitamin $B$ while food $Q$ contains $4$ $units/kg$ of Vitamin $A$ and $2$ $units/kg$ of vitamin $B$. Determine the minimum cost of the mixture. Let the mixture contain $x$ kg of food $P$ and $y$ kg of food $Q$. Therefore,$x \geq 0$ and $y \geq 0$. The given information can be compiled in a table as follows:

(A) Let the mixture contain $x$ kg of food $P$ and $y$ kg of food $Q$. Therefore,$x \geq 0$ and $y \geq 0$. The given information is compiled in the table below:

Food	Vitamin $A$ (units/kg)	Vitamin $B$ (units/kg)	Cost (Rs/kg)
$P$	$3$	$5$	$60$
$Q$	$4$	$2$	$80$
Requirement	$8$	$11$	-

The constraints are:
$3x + 4y \geq 8$
$5x + 2y \geq 11$
$x, y \geq 0$
Objective function to minimize: $Z = 60x + 80y$.
The corner points of the feasible region are $A(\frac{8}{3}, 0)$,$B(2, \frac{1}{2})$,and $C(0, \frac{11}{2})$.
Evaluating $Z$ at corner points:

Corner Point	$Z = 60x + 80y$
$A(\frac{8}{3}, 0)$	$60(\frac{8}{3}) + 80(0) = 160$
$B(2, \frac{1}{2})$	$60(2) + 80(\frac{1}{2}) = 120 + 40 = 160$
$C(0, \frac{11}{2})$	$60(0) + 80(\frac{11}{2}) = 440$

Since the feasible region is unbounded,we check if $60x + 80y < 160$ has any common points with the feasible region. The line $3x + 4y < 8$ does not intersect the feasible region. Thus,the minimum cost is Rs $160$ at any point on the line segment joining $A$ and $B$.