Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs. 60/kg and Food Q costs Rs. 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Answer & explanation
Correct answer: option 2
Let the mixture contain x kg of food P and y kg of food Q. Therefore,
$x≥0$ and $y≥0$
The given information can be compiled in a table as follows.
|
|
Vitamin A (units/kg) |
Vitamin B (units/kg) |
Cost (Rs/kg) |
|
Food P |
3 |
5 |
60 |
|
Food Q |
4 |
2 |
80 |
|
Requirement (units/kg) |
8 |
11 |
|
The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B
Therefore, the constraints are
$3x+4y≥8$
$5x+2y≥11$
Total cost, Z of purchasing food is, Z=60x+80y
The mathematical formulation of the given problem is
Minimise Z = $60x+80y$........(1)
subject to the constraints
$3x+4y≥8$......(2)
$5x+2y≥11$....(3)
$x,y≥0$......(4)
The feasible region determined by the system of constraints is as follows.
|
Corner point |
Z=60x+80y |
|
|
$A(\frac{8}{3},0)$ |
160 |
→ Minimum |
|
$B(2,\frac{1}{2})$ |
160 |
→ Minimum |
|
$C(0,\frac{11}{2})$ |
440 |
As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this, we graph the inequality, $60x+80y<160$ or $3x+4y<8$ 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 $3x+4y<8$
Therefore, the minimum cost of the mixture will be Rs.160 at the line segment joining the points $(\frac{8}{3},0)$ and $(2,\frac{1}{2})$
