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 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 and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements
Solution
Let the dietician wishes to mix x units of food F_{1} and y units of food F_{2}.
Clearly, \[x, y \geq 0\]
The given information can be tabulated as follows:
Vitamin A  Minerals  
Food F_{1}  3  4 
Food F_{2}  6  3 
Minimum requirement  80  100 
The constraints are
\[3x + 6y \geq 80\]
\[4x + 3y \geq 100\]
It is given that cost of food F_{1}_{ }and F_{2} is Rs 4 and Rs 6 per unit respectively. Therefore, cost of x units of food F_{1} and y units of food F_{2} 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
\[Z = 4x + 6y\] subject to
\[3x + 6y \geq 80\]
\[4x + 3y \geq 100\]
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
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
Corner point 
Z= 4x + 6y 
D \[\left( 0, \frac{100}{3} \right)\]  200 
E \[\left( 24, \frac{4}{3} \right)\]  104 
\[A\left( \frac{80}{3}, 0 \right)\]

\[\frac{320}{3}\]

The minimum value of Z is Rs 104 which is at E \[\left( 24, \frac{4}{3} \right)\]