A diet is to contain atleast 80 units of vitamin A and 100 units of minerals. Two foods $F_{1} a n d F_{2}$ are available. Food $F_{1}$ costs Rs. 4 per unit and food $F_{2}$ costs Rs. 6 per unit. One units of food $F_{1}$ contains at 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 diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

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Solution

Let the diet contains x unit of food $F_{1}$ and y units of food $F_{2}$ .

We construct the following table :

$\begin{matrix} T y p e & N u m b e r & V i t a m i n A p e r & M i n e r a l s p e r u n i t & C o s t u n i t & (i n R s .) F_{1} & x & 3 x & 4 x & 4 x F_{2} & y & 6 y & 3 y & 6 y T o t a l & x + y & 3 x + 6 y & 4 x + 3 y & 4 x + 6 y \end{matrix}$

The cost of food $F_{1}$ is Rs. 4 per unit and of food $F_{2}$ is Rs. 6 per unit.

So, our problem is to minimize Z = 4x + 6y .....(i)

Subject to constraints $3 x + 6 y \geq 80$ ........(ii)

$4 x + 3 y \geq 100$ .........(iii)

$x \geq 0, y \geq 0$ .........(iv)

Firstly, draw the graph of the line 3x + 6y = 80

$\begin{matrix} x & 0 & \frac{80}{3} y & \frac{40}{3} & 0 \end{matrix}$

Putting (0, 0) in the inequality $3 x + 6 y \geq 80$ , we have $3 \times 0 + 6 \times 0 \geq 80 \Rightarrow 0 \geq 80$ (which is false)

So, the half plane is away from the origin.

Since, x, y $\geq$ 0

So, the feasible region lies in the first quadrant.

Secondly, draw the graph of the line $4 x + 3 y = 100$

$\begin{matrix} x & 0 & 25 y & \frac{100}{3} & 0 \end{matrix}$

Putting (0, 0) in the inequality 4x + 3y $\geq$ 100, we have

$4 \times 0 + 3 \times 0 \geq 100 \Rightarrow 0 \geq 100$ (which is false)

So, the half plane is away from the origin.

On solving the equations $3 x + 6 y = 80$ and 4x + 3y = 100, we get $B (24, \frac{4}{3})$

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are $A (\frac{80}{3}, 0), B (24, \frac{4}{3})$ and $(0, \frac{100}{3})$ . The values of Z at these points are as follows :

$\begin{matrix} C o r n e r p o i n t & Z = 4 x + 6 y A (\frac{80}{3}, 0) & \frac{320}{3} = 106.67 B (24, \frac{4}{3}) & 104 \to M i n i m u m C (0, \frac{100}{3}) & 200 \end{matrix}$

As the feasible is unbounded therefore, 104 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, $4 x + 6 y < 104 o r 2 x + 3 y < 52$ and check, whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with $2 x + 2 y < 52$

Therefore, the minimum cost of the mixture will be Rs. 104.

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A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁and F₂ are available. Food F₁ costs Rs 4 per unit food and F₂ costs Rs 6 per unit. One unit of food F₁contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ 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?