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 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 maximise the amount of vitamin A in the diet ? What is the maximum amount of vitamin A in the diet?

.: 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:

Maximize $Z = 6x + 3y$ (vitamin A)
Subject to the constraints

$12x + 3y \ge 240$ (constraint on calcium),

i.e. $4x + y \ge 80$ ...(1)

$4x + 20y \ge 460$ (constraint on iron), i.e., $x + 5y \ge 115$ ... (2)

$6x + 4y \le 300$ (constraint on cholesterol),

i.e., $3x + 2y \le 150$ …(3)

$x \ge 0,y \ge 0$ ….(4)

${l_1}:4x + y = 80$;

${l_2}:x + 5y = 115;{l_3}:3x + 2y = 150$

Let us graph the inequalities (1) to (4).

The feasible region (shaded)determined by the constraints

(1) to (4) is shown in graph 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.

From the table, we find that Z is maximum at the point (40, 15).

Hence, the amount of vitamin A under the constraints given in the problem will be maximum,

if 40 packets of food P and 15 packets of food Q are used in the special diet.

The maximum amount of vitamin A will be 285 units.