Question

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 4units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

Answer 1

Let the mixture contain x kg of food P and y kg of food Q.

Therefore, x≥0andy≥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
The total cost, Z, of purchasing food is Z=60x+80y

The mathematical foundation 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.

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(\(\frac{8}{3}\),0),B(2,\(\frac{1}{2}\)), and C(0,\(\frac{11}{2}\)) The values of Z at these corner points are 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 Rs160 at the line segment joining the points (\(\frac{8}{3}\),0)and(2,\(\frac{1}{2}\)).