Question

(Diet problem) A dietician has to develop a special diet using two foods $P$ and $Q$. Each packet (containing $30g$) of food $P$ contains $12$ units of calcium, $4$ units of iron, $6$ units of cholesterol and $6$ units of vitamin A. Each packet of the same quantity of food $Q$ contains $3$ units of calcium, $20$ units of iron, $4$ units of cholesterol and $3$ units of vitamin A. The diet requires atleast $240$ units of calcium, atleast $460$ units of iron and at most $300$ units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?

Answer 1

Let $x$ and $y$ be the number of packets of food $P$ and $Q$ respectively. Obviously $x\ge 0,y\ge 0$. Mathematical formulation of the given problem is as follows :
Minimise $Z=6x+3y$ (vitamin A)
subject to the constraints
$12x+3y\ge 240$ (constraint on calcium), i.e. $4x+y\ge 80...\left(1\right)$
$4x+20y\ge 460$ (constraint on iron), i.e. $x+5y\ge 115...\left(2\right)$
$6x+4y\le 300$ (constraint on cholesterol), i.e. $3x+2y\le 150...\left(3\right)$
$x\ge 0,y\ge 0...\left(4\right)$
Let us graph the inequalities $\left(1\right)$ to $\left(4\right)$.
The feasible region (shaded) determined by the constraints $\left(1\right)$ to $\left(4\right)$ is shown in Fig and note that it is bounded.
The coordinates of the corner points $L,M$ and $N$ are $(2,72),(15,20)$ and $(40,15)$ respectively. Let us evaluate $Z$ at these points:

Corner Point	$Z=6x+3y$
$(2,72)$	$228$
$(15,20)$	$150\leftarrow Minimum$
$(40,15)$	$285$

From the table, we find that $Z$ is minimum at the point $(15,20)$. Hence, the amount of vitamin A under the constraints given in the problem will be minimum, if $15$ packets of food $P$ and $20$ packets of food $Q$ are used in the special diet. The minimum amount of vitamin A will be $150$ units.