Question

A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:

	Vitamin A	Vitamin B	Vitamin C	Vitamin D
Food X Food Y	1 2	1 1	1 3	2 1

One kg food X costs Rs 5, whereas one kg of food Y costs Rs 8. Find the least cost of the mixture which will produce the desired diet.

Answer 1

Let the dietician wishes to mix x kg of food X and y kg of food Y.

Therefore, $x,y\ge 0$
As we are given,

	Vitamin A	Vitamin B	Vitamin C	Vitamin D
Food X Food Y	1 2	1 1	1 3	2 1

It is given that the mixture should contain at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D.

Therefore, the constraints are
$$\begin{gathered} x+2y\ge 6 \\ x+y\ge 7 \\ x+3y\ge 11 \\ 2x+y\ge 9 \end{gathered}$$

It is given that cost of food X is Rs 5 per kg and cost of food Y is Rs 8 per kg.

Thus, Z = $5x+8y$

Thus, the mathematical formulat​ion of the given linear programmimg problem is

Minimize Z = $5x+8y$

subject to

$$\begin{gathered} x+2y\ge 6 \\ x+y\ge 7 \\ x+3y\ge 11 \\ 2x+y\ge 9 \end{gathered}$$

First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 6, x + y = 7, x + 3y =11, 2x + y =9, x = 0 and y = 0.

The line x + 2y = 6 meets the coordinate axis at A₁(6, 0) and B₁(0, 3). Join these points to obtain the line x + 2y = 6. Clearly, (0, 0) does not satisfies the inequation x + 2y ≥ 6. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line x + y = 7 meets the coordinate axis at C₁(7, 0) and D₁(0, 7). Join these points to obtain the line x + y = 7. Clearly, (0, 0) does not satisfies the inequation x + y ≥ 7. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line x + 3y = 11 meets the coordinate axis at $E_1\left(11,0\right)$ and $F_1\left(0,\frac{11}{3}\right)$. Join these points to obtain the line x + 3y = 11.Clearly, (0, 0) does not satisfies the inequation x + 3y ≥ 11. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line 2x + y = 9 meets the coordinate axis at $G_1\left(\frac{9}{2},0\right)$ and $H_1\left(0,9\right)$. Join these points to obtain the line 2x + y = 9.Clearly, (0, 0) does not satisfies the inequation 2x + y ≥ 9. 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 H₁(0, 9), I₁(2 ,5), J₁(5, 2), E₁(11, 0).

The values of Z at these corner points are as follows

Corner point	Z= 5x + 8y
H1	72
I1	50
J1	41
E1	55

The minimum value of Z is at J₁(5, 2) which is Rs 41.

Hence, cheapest combination of foods will be 5 units of food X and 2 units of food Y.