$A$ dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain at least $8$ $units$ of vitamin $A$ and $10$ $units$ of vitamin $C$. Food $'I'$ contains $2$ $units/kg$ of vitamin $A$ and $1$ $unit/kg$ of vitamin $C$. Food $'II'$ contains $1$ $unit/kg$ of vitamin $A$ and $2$ $units/kg$ of vitamin $C$. It costs Rs $50$ per $kg$ to purchase Food $'I'$ and Rs $70$ per $kg$ to purchase Food $'II'$. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.

(B) Let the mixture contain $x$ $kg$ of Food $'I'$ and $y$ $kg$ of Food $'II'$. Clearly,$x \geq 0, y \geq 0$.
We make the following table from the given data:

Resources	Food $I$ $(x)$	Food $II$ $(y)$	Minimum Requirement
Vitamin $A$ (units/kg)	$2$	$1$	$8$
Vitamin $C$ (units/kg)	$1$	$2$	$10$
Cost (Rs/kg)	$50$	$70$	Minimize $Z$

Since the mixture must contain at least $8$ units of vitamin $A$ and $10$ units of vitamin $C$,we have the constraints:
$2x + y \geq 8$
$x + 2y \geq 10$
Total cost $Z$ of purchasing $x$ $kg$ of food $'I'$ and $y$ $kg$ of Food $'II'$ is $Z = 50x + 70y$.
Hence,the mathematical formulation of the problem is:
Minimize $Z = 50x + 70y$ subject to the constraints:
$2x + y \geq 8$
$x + 2y \geq 10$
$x, y \geq 0$
Let us graph the inequalities. The feasible region is unbounded. Let us evaluate $Z$ at the corner points $A(0, 8)$,$B(2, 4)$,and $C(10, 0)$.

Corner point	$Z = 50x + 70y$
$(0, 8)$	$560$
$(2, 4)$	$380$ (Minimum)
$(10, 0)$	$500$

The smallest value of $Z$ is $380$ at the point $(2, 4)$. Since the feasible region is unbounded,we check the inequality $50x + 70y < 380$ or $5x + 7y < 38$. As this region has no common points with the feasible region,the minimum value is indeed $380$.
Thus,the optimal mixing strategy is to mix $2$ $kg$ of Food $'I'$ and $4$ $kg$ of Food $'II'$,with a minimum cost of Rs $380$.