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 $Rs.4$ per unit food and $F_2$ costs $Rs.6$ per unit. One unit of food $F_1$ contains $3$ units of vitamin A and $4$ units of minerals. One unit of food $F_2$ contains $6$ units of vitamin A and $3$ units of minerals. Formulate this as a linear programming problem. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.

Answer 1

Let's assume that diet contains $Xunits$ of food $F_1$ and $Yunits$ of food $F_2$.

Cost of 1 unit of food $F_1=4Rs$

Cost of 1 unit of food $F_2=6Rs$

So, Total cost (C) $=4X+6YRs...\left(1\right)$

Now, food $F_1$ contains $3units$ of vitamin A and food $F_2$ contains $6units$ of vitamin A.

So, Total Vitamin A contains $=3X+6Y$

Since, minimum requirement of Vitamin A is 80 units.

$\therefore 3X+6Y\ge 80...\left(2\right)$

Also, food $F_1$ contains $4units$ of minerals and food $F_2$ contains $3units$ of minerals.

So, Total minerals contains $=4X+3Y$

Since, minimum requirement of minerals is 100 units.

$\therefore 4X+3Y\ge 100...\left(3\right)$

Since, amount of food can never be negative.

So, $X\ge 0,Y\ge 0...\left(4\right)$

We have to minimise the cost of diet given in equation (1) subject to the constraints given in (2), (3) and (4).

After plotting all the constraints, we get the feasible region as shown in the image.

Corner points	Value of $C=4X+6Y$
A $(0,\frac{100}{3})$	$200$
B $(24,\frac{4}{3})$	$104$ (Minimum)
C $(\frac{80}{3},0)$	$106\frac{2}{3}$

Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, plot the region $Z<104$ to check if there exist some points in feasible region where value can be less than 160.

$\Rightarrow 4X+6Y<104$

Since there is no common points between the feasible region and the region which contains $Z<104$. So 104 is the minimum cost.