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 and 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 diet that consists of mixture, of these two foods and also meets the minimal nutritional requirements.

.: Let the mixture contain x kg of food ${F_1}$ and y kg of food ${F_2}.$ Clearly,

$x \ge 0,y \ge 0$. We make the following table from the given data :

Since the mixture contains atleast 80 units of vitamin A and 100 units of minerals,

we have the constraints
$3x + 6y \ge 80;$

$4x + 3y \ge 100$
Total cost Z of purchasing x units of food ${F_1}$ and y units of food ${F_2}$ is $Z = 4x + 6y$

Hence, the mathematical formulation of the problem is

Minimize $Z = 4x + 6y$ ...(1)

subject to the constraints
$3x + 6y \ge 80$ ...(2)

$4x + 3y \ge 100$ ...(3)

$x,y \ge 0$ ...(4)

Let ${l_1}:3x + 6y = 80,$

${l_2}:4x + 3y = 100$

Let us graph the inequalities (2) to (4).

The feasible region determined by the system is shown in the graph.

Here, observe that the feasible region is unbounded.

Let us evaluate Z at the comer points
$A\left( {\cfrac{{80}}{3},0} \right),E\left( {24,\cfrac{4}{3}} \right),D\left( {0,\cfrac{{100}}{3}} \right)$

In the table, we find that the smallest value of Z is 104 at the point E(24, 4/3).

Since the feasible region is unbounded, therefore we have to draw the graph of the

inequality $4x + 6y < 104$ i.e.$2x + 3y < 52$ to check whether

the resulting open half plane has any point common with the feasible region.

From the graph, we see that it has no points in common.

Thus, the minimum value of Z is 104 attained at the point (24, 4/3).

Therefore, the minimum cost for diet is Rs.104.