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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?

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Let the dietician mixes x kg of food X and y kg of food Y. We construct the following table: 

Food Amount Vitamin A Vitamin B Vitamin C Cost
X x kg x 2x 3x 16x
Y y kg 2y 2y y 20y
Total (x + y)kg x + 2y 2x + 2y 3x + y 16x + 20y
Minimum Requires 10 12 8

So, our problem is to minimize Z = 16x + 20y …(i) 

Subject to constraints x + 2y ≥ 10 …(ii) 

2x + 2y ≥ 12 x + y ≥ 6 …(iii) 

3x + y ≥ 8 …(iv) 

x ≥ 0, y ≥ 0 …(v) 

Firstly, draw the graph of the line x + 2y = 10 

Secondly, draw the graph of the line line x + y = 6 

Thirdly, draw the graph of the line 3x + y = 8 

On solving equations x + y = 6 and x + 2y = 10, we get B(2, 4) 

Similarly, solving the equations 3x + y = 8 and x + y = 6, we get C(1, 5). 

The corner points of the feasible region are A(10, 0), B(2, 4), C(1, 5) and D(0, 8).

The values of Z at these points are as follows:

As the feasible region is unbounded, therefore 112 may or may not be the minimum value of Z. 

For this,we drawa graph of the inequality,16x + 20y < 112 or4x + 5y < 28 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 4x + 5y < 28. 

Therefore, the minimum value of Z is 112 at B(2, 4). 

Thus, the mixture should contain 2 kg of food X and 4 kg of food Y. The minimum cost of the mixture is Rs. 112.

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