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 diet that consists of mixture of these two foods and also meets the minimal nutritional requirements?
Solution
Let the diet contain x units of food F_{1} and y units of food F_{2}. Therefore,
x ≥ 0 and y ≥ 0
The given information can be complied in a table as follows.
Vitamin A (units)  Mineral (units) 
Cost per unit (Rs) 

Food F_{1} (x)  3  4  4 
Food F_{2} (y)  6  3  6 
Requirement  80  100 
The cost of food F_{1} is Rs 4 per unit and of Food F_{2} is Rs 6 per unit. Therefore, the constraints are
3x + 6y ≥ 80
4x + 3y ≥ 100
x, y ≥ 0
Total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem is
Minimise Z = 4x + 6y … (1)
subject to the constraints,
3x + 6y ≥ 80 … (2)
4x + 3y ≥ 100 … (3)
x, y ≥ 0 … (4)
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(8/3,0), B(2, 1/2), and C(0, 11/2).
The corner points are A(8/3,0), B(24,4/3) and C(0, 100/3).
The values of Z at these corner points are as follows.
Corner point  Z = 4x + 6y  
A(8/3,0)  320/3 = 106.67  
B(24,4/3)  104  → Minimum 
C(0, 100/3)  200 
As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.
For this, we draw a graph of 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