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.