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 content of one kg of 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 & explanation
Correct answer: option 3
Let the mixture contain x kg of food X and y kg of food Y
The mathematical formulation of the given problem is as follows.
Minimize $z=16x+20y$......(1)
subject to the constraints,
$x+2y≥10$......(2)
$x+y≥6$.........(3)
$3x+y≥8$.......(4)
$x,y≥0$..........(5)
The feasible region determined by the system of constraints is as follows:
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 corner points are as follows.
|
Corner point |
z=16x+20y |
|
|
A(10,0) |
160 |
|
|
B(2,4) |
112 |
→ Minimum |
|
C(1,5) |
116 |
|
|
D(0,8) |
160 |
As the feasible region is unbounded, therefore, 112 may or may not be the minimum value of z
For this, we draw a graph of the inequality, $16x+20y<112$ or $4x+5y<28$ 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 (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
