Question

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

• A. ₹ 160
• B. ₹ 150
• C. ₹ 130
• D. ₹ 140

Answer 1

Answer: (A)

Let Reshma mixes $xkg$ of food $P$ and $ykg$ 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 $Z=60x+80y$
The mathematical formulation of the given problem is
Minimise $Z=60x+80y$....(i)
Subject to the constraints $3x+4x\ge 8$...(ii)
$5x+2y\ge 11$...(iii)
$x\ge 0,y\ge 0$....(iv)
Draw the graph of the line $3x+4y=8$.

x	0	8/3
y	2	0

Putting $(0,0)$ in the inequality $3x+4y\ge 8$, we have
$3\times 0+4\times 0\ge 8$
$0\ge 8\text{ (which is false) }$
So, the hali plane is away from the origin.
Since, $x,y\ge 0$
So, the feasible region lies in the first quadrant.
Draw the graph of the line $5x+2y=11$.

x	0	11/5
y	11/2	0

Putting $(0,0)$ in the inequality $5x+2y\ge 11$, we have
$5\times 0+2\times 0\ge 11$
$\to 0\ge 11$ (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 $3x+4y-8$ and $5x+2y=11$, 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	$z=60x+80y$
$A\left(\frac{8}{3},0\right)$	$160\to$ Minimum
$B\left(2,\frac{1}{2}\right)$	$160\to$ 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 $60x+80y<160$ 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 $₹160$ at line segment joining the points $A\left(\frac{8}{3},0\right)$ and $B\left(2,\frac{1}{2}\right)$.