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 minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?
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. Clearly x ≥ 0, y ≥ 0.
Let Z be the quantity used of vitamin A.
Mathematical formulation of the given problem is as follows:
Minimise    Z = 6x + 3y
subject to the constraints
12x + 3y ≥ 240 i.e. 4 x + y ≥ 80
4x + 20y ≥ 460 i.e. x + 5 y ≥ 115
6x + 4y ≤ 300 i.e. 3 x + 2 y ≤ 150
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 4x + y = 80
For x = 0, y = 80
For y = 0, 4 x = 80 or x = 20
∴ line meets OX in A(20, 0) and OY in L(0, 80).
Also we draw the graph of x + 5y = 115
For x = 0, 5y = 115 or y = 23
For y = 0, x = 115
∴ line meets OX in E (115, 0) and OY in M(0, 23)
Again consider
3x + 2y = 150
For x = 0, 2 y = 150 or y = 75
For y = 0, 3 x = 150 or x = 50
∴ line meets OX in C(50, 0) and OY in N (0, 75).
Since feasible region satisfies all the constraints.
∴ DEF is the feasible region.

The corner points are D(40, 15), E(15, 20), F(2, 72).
At D(40, 15), Z = 6 × 40 + 3 × 15 = 240 + 45 = 285
At E(15, 20), Z = 6 × 15 + 3 × 20 = 90 + 60 = 150
At F(2, 72), Z = 6 × 2 + 3 × 72 = 12 + 216 = 228
∴ minimum value = 150 at (15, 20).
∴ 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 speciai diet. The minimum amount of vitamin A will be 150 units.
Now instead of minimising Z, we have to maximise Z.
∴ maximum value = 285 at (40, 15)
∴ the amount of vitamin A under the constraints given in the problem will be maximum, if 40 packets of focd P and 15 packets of food Q are used in the special diet. The maximum amount of vitamin A will be 285 units.