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

Answer 1

Let x units of type P and y units of type Q of food be mixed.

$\mathrm{Clearly},x,y\ge 0$

The given information can be tabulated as follows:

Food	Vitamin A(kg)	Vitamin B(kg)
P	3	5
Q	4	2
Requirement	8	11

Therefore, the constraints are
$$\begin{gathered} 3x+4y\ge 8 \\ 5x+2y\ge 11 \end{gathered}$$

Food P costs Rs 60 kg and Food Q costs Rs 80 kg. Therefore, cost of x units of type P and y units of type Q of food be Rs 60x and Rs 80y respectively.
Total cost = Z = 60x + 80y

​Thus, the mathematical formulation of the given LPP is

Min Z = 60x + 80y
subject to

$$\begin{gathered} 3x+4y\ge 8 \\ 5x+2y\ge 11 \end{gathered}$$

First we will convert inequations into equations as follows:9
3x + 4y = 8, 5x + 2y = 11, x = 0 and y = 0

Region represented by 3x + 4y ≥ 8:
The line 3x + 4y = 8 meets the coordinate axes at $A\left(\frac{8}{3},0\right)$ and $B\left(0,2\right)$ respectively. By joining these points we obtain the line
3x + 4y = 8. Clearly (0,0) does not satisfies the 3x + 4y = 8. So, the region which does not contain the origin represents the solution set of the inequation 3x + 4y ≥ 8.

Region represented by 5x + 2y ≥ 11:
The line 5x + 2y = 11 meets the coordinate axes at $C\left(\frac{11}{5},0\right)$ and $D\left(0,\frac{11}{2}\right)$ respectively. By joining these points we obtain the line
5x + 2y = 11. Clearly (0,0) satisfies the inequation 5x + 2y ≥ 11. So,the region which contains the origin represents the solution set of the inequation 5x + 2y ≥ 11.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 3x + 4y ≥ 8, 5x + 2y ≥ 11, x ≥ 0 and y ≥ 0 are as follows.




The corner points are $D\left(0,\frac{11}{2}\right)$,$E\left(2,\frac{1}{2}\right)$, $A\left(\frac{8}{3},0\right)$

The value of the objective function at the corner points

Corner points	Z = 60x + 80y
$D\left(0,\frac{11}{2}\right)$	440
$E\left(2,\frac{1}{2}\right)$	160
$A\left(\frac{8}{3},0\right)$	160

The minimum value of Z is 160 which is attained at $E\left(2,\frac{1}{2}\right)$ and $A\left(\frac{8}{3},0\right)$.

Thus, the minimum cost is Rs 160 obtained when 2 kg of food P and 0.5 kg of food Q is mixed or when 2.67 kg of food P and 0 kg of food Q is mixed.