Question

A diet is to contain at least $80$ units of vitamin $A$ and $100$ units of minerals. Two foods $F_1$ and $F_2$ are available. Food $F_1$ costs $\text{Rs}4$ per unit and food $F_2$ costs $\text{Rs}6$ per unit. One unit of food $F_1$ contains $2$ units of vitamin $A$ and $4$ units of minerals and one unit of $F_2$ contains $4$ units of vitamin $A$ and $3$ units of minerals. Then for minimum cost of diet which will produce the required diet, which of the following is correct?

• A. The minimum cost of diet is $\text{Rs}136$
• B. quantity of food $F_1$ is $16$ units
• C. quantity of food $F_1$ is $12$ units
• D. optimal solution is not possible.

Answer 1

Answer: (A), (B)

quantity of food $F_1$ is $16$ units

Let a mixture of $x$ amount of $F_1$ and $y$ amount of $F_2$ has been taken into a diet.
Objective function : $Z=4x+6y$
Constraints :
For Vitamin $A,$
$2x+4y\ge 80\Rightarrow x+2y\ge 40\cdots \left(i\right)$
For minerals, $4x+3y\ge 100\cdots \left(ii\right)$
and $x,y\ge 0\cdots \left(iii\right)$
Plotting the graph of constraints :

We can see feasible region is unbounded.
Now tabulating the value of $Z=4x+6y$ at corner points :

Corner points :$(x,y)$	Value : $Z=4x+6y$
$A(0,\frac{100}{3})$	$200$
$E(16,12)$	$136$
$D(40,0)$	$160$

From above table, we get $Z_{min}=136,$ but feasible region is unbounded, so we have to check if any other minimum value is possible.
Now, plotting the curve, $4x+6y<136,$ if no other value is possible.

From graph, we can see that only point $E(16,12)$ is common for $4x+6y<136$ in the feasible unbounded region.
So, $Z_{min}=136$