Question

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 $A\text{and}10$ units of vitamin $C$. Food ${}^{\prime }{I}^{\prime }$ contains $2\text{units/kg}$ of vitamin $A$ and $1\text{unit/kg}$ of vitamin $C$. Food ${}^{\prime }I{I}^{\prime }$ contains $1\text{unit/kg}$ of vitamin $A$ and $2\text{unit/kg}$ of vitamin $C$. It costs $Rs.50\text{per kg}$ to purchase Food ${}^{\prime }{I}^{\prime }$ and $Rs.70\text{per kg}$ to purchase Food ${}^{\prime }I{I}^{\prime }$. Formulate this problem as a linear programming problem to minimize the cost of such a mixture.

Answer 1

Let the mixture contains $x$ kg of food $I$ and $y$ kg of food $II$

Combining the constraints:
Minimize $Z=50x+70y$
Subject to constraints:
$2x+y\ge 8$
$x+2y\ge 10$
$x,y\ge 0$

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

$x+2y=10$
$\begin{array}{ccc}x& 0& 10\\ y& 5& 0\end{array}$
$\begin{array}{cc}\text{Corner Points}& \text{Value of}Z=50x+70y\\ (0,8)& 560\\ (2,4)& 380\\ (10,0)& 500\end{array}$

Since the feasible region is unbounded.
Hence $380$ may or may not be the minimu value of $Z$
$50x+70y<380$
$50x+70y=380$
$\begin{array}{ccc}x& 0& \frac{38}{5}\\ y& \frac{38}{7}& 0\end{array}$
As there is no common points between the feasible region and inequality.
$\therefore 380$ is the minimum value of $Z$.
Therefore, Minimum cost is $Rs.380$ when $2$ kg of food ${}^{\prime }{I}^{\prime }$ is mixed with $4$ kg of food ${}^{\prime }I{I}^{\prime }$.