Q. Reshma wishes to mix two types of food $P$ and $Q$ in such a way that the vitamin contents of the mixture contain atleast 8 units of vitamin $A$ and 11 units of vitamin B. Food $P$ costs $₹ 60 / kg$ and food $Q$ costs ₹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. Then, the minimum cost of the mixture is
Linear Programming
Solution:
Let Reshma mixes $x kg$ of food $P$ and $y kg$ of food $Q$. Construct the following table
Food
Quantity
Vitamin A
Vitamin B
(₹ per kg)
P
x kg
3x
5x
60x
Q
y kg
4y
2y
80y
Total
3 x + 4 y
5 x +2 y
60x+80y
Requirement
3 4 x y
Atleast 11
The mixture must contain atleast 8 units of vitamin $A$ and 11 units of vitamin B.
Total cost $Z$ of purchasing food is $Z=60 x+80 y$
The mathematical formulation of the given problem is
Minimise $Z=60 x+80 y$....(i)
Subject to the constraints $3 x+4 x \geq 8$...(ii)
$5 x+2 y \geq 11 $...(iii)
$ x \geq 0, y \geq 0$....(iv)
Draw the graph of the line $3 x+4 y=8$.
x
0
8/3
y
2
0
Putting $(0,0)$ in the inequality $3 x+4 y \geq 8$, we have
$3 \times 0+4 \times 0 \geq 8$
$0 \geq 8 \quad \text { (which is false) }$
So, the hali plane is away from the origin.
Since, $x, y \geq 0$
So, the feasible region lies in the first quadrant.
Draw the graph of the line $5 x+2 y=11$.
x
0
11/5
y
11/2
0
Putting $(0,0)$ in the inequality $5 x+2 y \geq 11$, we have
$5 \times 0+2 \times 0 \geq 11$
$\rightarrow 0 \geq 11$ (which is false)
So, the half plane is away from the origin.
It can be seen that the feasible region is unbounded.
On solving equations $3 x+4 y-8$ and $5 x+2 y=11$, we get $B\left(2, \frac{1}{2}\right)$.
The corner points of the feasible region are $A\left(\frac{8}{3}, 0\right)$, $B\left(2, \frac{1}{2}\right)$ and $C\left(0, \frac{11}{2}\right)$.
The values of $Z$ at these points are as follows
Corner point
$z=60 x+80 y$
$A\left(\frac{8}{3}, 0\right)$
$160 \rightarrow$ Minimum
$B\left(2, \frac{1}{2}\right)$
$160 \rightarrow$ Minimum
$C\left(0, \frac{11}{2}\right)$
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 $60 x+80 y<160$ or $3 x+4 y<8$ 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 $3 x+4 y<8$, therefore the minimum cost of the mixture will be $₹ 160$ at line segment joining the points $A\left(\frac{8}{3}, 0\right)$ and $B\left(2, \frac{1}{2}\right)$.
| Food | Quantity | Vitamin A | Vitamin B | (₹ per kg) |
|---|---|---|---|---|
| P | x kg | 3x | 5x | 60x |
| Q | y kg | 4y | 2y | 80y |
| Total | 3 x + 4 y | 5 x +2 y | 60x+80y | |
| Requirement | 3 4 x y | Atleast 11 |
| x | 0 | 8/3 |
| y | 2 | 0 |
| x | 0 | 11/5 |
| y | 11/2 | 0 |
| Corner point | $z=60 x+80 y$ |
|---|---|
| $A\left(\frac{8}{3}, 0\right)$ | $160 \rightarrow$ Minimum |
| $B\left(2, \frac{1}{2}\right)$ | $160 \rightarrow$ Minimum |
| $C\left(0, \frac{11}{2}\right)$ | 440 |