A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs Rs 4 per unit and F₂ costs Rs 6 per unit one unit of food F₁ contains 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 and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements.

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Solution

Let the dietician wishes to mix x units of food F₁ and y units of food F₂.
Clearly, $x, y \geq 0$

The given information can be tabulated as follows:

Vitamin A Minerals

Food F₁ 3 4

Food F₂ 6 3

Minimum requirement 80 100

The constraints are
$3 x + 6 y \geq 80 4 x + 3 y \geq 100$

It is given that cost of food F₁and F₂ is Rs 4 and Rs 6 per unit respectively. Therefore, cost of x units of food F₁ and y units of food F₂ is Rs 4x and Rs 6y respectively.
Let Z denote the total cost
∴ Z = 4x + 6y

Thus, the mathematical formulation of the given linear programmimg problem is
Minimize $Z = 4 x + 6 y$ subject to

$3 x + 6 y \geq 80 4 x + 3 y \geq 100$
$x, y \geq 0$

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 6y = 80, 4x + 3y = 100, x = 0 and y = 0

The line 3x + 6y = 80 meets the coordinate axis at $A (\frac{80}{3}, 0)$ and $B (0, \frac{40}{3})$ . Join these points to obtain the line 3x + 6y = 80. Clearly, (0, 0) does not satisfies the inequation 3x + 6y ≥ 80. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line 4x + 3y = 100 meets the coordinate axis at C(25, 0) and $D (0, \frac{100}{3})$ . Join these points to obtain the line 4x + 3y = 100. Clearly, (0, 0) does not satisfies the inequation 4x + 3y ≥ 100. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The feasible region determined by the system of constraints is

The corner points are D $(0, \frac{100}{3})$ , E $(24, \frac{4}{3})$ and $A (\frac{80}{3}, 0)$ .

The values of Z at these corner points are as follows:

Corner point Z= 4x + 6y

D $(0, \frac{100}{3})$ 200

E $(24, \frac{4}{3})$ 104

$A (\frac{80}{3}, 0)$ $\frac{320}{3}$

The minimum value of Z is Rs 104 which is at E $(24, \frac{4}{3})$ .

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