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.

160 

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.

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.

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 $(2,\frac{1}{2})$