Question

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 Rs4 per unit food and F₂ costs Rs6 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. 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 the diet contain
x units of food F₁ and y units of food F₂.

Therefore, x≥0 and y≥0

The given information can be compiled in a table as follows.

	Vitamin A(units)	Mineral(units)	Cost per unit (Rs)
Food F1(x)	3	4	4
Food F2(y)	6	3	6
Requirement	80	100

Therefore, the constraints are 3x+6y≥80 4x+3y≥100 x,y≥0

The feasible region determined by the constraints is as follows.

It can be seen that the feasible region is unbounded. The corner points of the feasible region are A(\(\frac{8}{3}\),0), B(2,\(\frac{1}{2}\)) and C(0,\(\frac{11}{2}\))

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

As the feasible region is unbounded,

therefore, 104 may or may not be the minimum value of Z.
For this, we draw a graph for the inequality, 4x+6y<104 or 2x+3y<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 2x+3y<52

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