Question

A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) 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 at least 240 units of calcium, at least 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

Answer 1

Simplification of given data
Let Number of Packets of food P be 𝑥 Number of Packets of food Q be 𝑦 According to Question:

Resources	Foods	Requirement
P	Q
Calcium	12 units	3 units	at least 240 units
Iron	4 units	20 units	at least 460 units
Cholesterol	6 units	4 units	at most 300 units
Vitamin A	6 units	3 units

Combining all the constraints:
maximise
Z=6x+3y
Subject to Constraints,
$12x+3y\ge 240\Rightarrow 4x+y\ge 80$
$4x+20y\ge 460\Rightarrow x+5y\ge 115,$
$6x+4y\le 300\Rightarrow 3x+2y\le 150,$
$x\ge 0,y\ge 0$
Given graph
4x+y =80
$\begin{array}{ccc}x& 0& 20\\ y& 80& 0\end{array}$

x+5y =115
$\begin{array}{ccc}x& 0& 70\\ y& 23& 9\end{array}$

3x+2y =150
$\begin{array}{ccc}x& 0& 50\\ y& 75& 0\end{array}$

Find maximum amount of Vitamin A
$\begin{array}{cc}\text{Corner Points}& \text{value of}Z=6x-3y\\ (2,72)& 228\\ (15,20)& 150\\ (40,15)& 285\end{array}$
Hence, the amount of Vitamin A will be maximum if 40 Packets of Food P & 15 Packets of Food Q are used.
Thus, maximum amount of Vitamin A =285 units.