Question

A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

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

One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?

Answer 1

Let the dietician wishes to mix x kg of food X and y kg of food Y.
Quantity of food cannot be negative.
Therefore, $x,y\ge 0$

It is given that the mixture should contain at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C.
The vitamin contents of one kg food is given below:

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

Therefore, the constraints are

$$\begin{gathered} x+2y\ge 10 \\ 2x+2y\ge 12 \\ 3x+y\ge 8 \end{gathered}$$

It is given that cost of food X is Rs 16 per kg and cost of food Y is Rs 20 per kg. Therefore, cost of x kg of food X and y kg of food Y is Rs 16x and Rs 20y respectively.
Thus, total cost Z = $16x+20y$

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

Minimize Z = $16x+20y$

subject to

$$\begin{gathered} x+2y\ge 10 \\ 2x+2y\ge 12 \\ 3x+y\ge 8 \end{gathered}$$

First we will convert inequations into equations as follows:
x + 2y = 10, 2x + 2y = 12, 3x + y = 8, x = 0 and y = 0

Region represented by x + 2y ≥ 10:
The line x + 2y = 10 meets the coordinate axes at A₁(10, 0) and B₁(0, 5) respectively. By joining these points we obtain the line x + 2y = 10 . Clearly (0,0) does not satisfies the x + 2y = 10. So, the region which does not contain the origin represents the solution set of the inequation x + 2y ≥ 10.

Region represented by 2x + 2y ≥ 12:
The line 2x + 2y = 12 meets the coordinate axes at C₁(6, 0) and D₁(0, 6) respectively. By joining these points we obtain the line 2x + 2y = 12. Clearly (0,0) does not satisfies the inequation 2x + 2y ≥ 12. So,the region which does not contain the origin represents the solution set of the inequation 2x + 2y ≥ 12.

Region represented by 3x + y ≥ 8:
The line 3x + y = 8 meets the coordinate axes at $E_1\left(\frac{8}{3},0\right)$ and F₁(0, 8) respectively. By joining these points we obtain the line 3x + y = 8 . Clearly (0,0) does not satisfies the inequation 3x + y ≥ 8. So,the region which does not contain the origin represents the solution set of the inequation 3x + y ≥ 8.

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 x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints x + 2y ≥ 10, 2x + 2y ≥ 12, 3x + y ≥ 8, x ≥ 0 and y ≥ 0 are as follows:



The corner points are F₁(0, 8), G₁(1, 5), H₁(2, 4) and A₁(10, 0).

The values of Z at these corner points are as follows

Corner point	Z= 16x + 20y
F1	160
G1	116
H1	112
A1	160

The minimum value of Z is at H₁(2, 4) which is 112.
Hence, cheapest combination of foods will be 2 kg of food X and 4 kg of food Y and the minimum cost is Rs 112.