Question

P and Q are two food packets. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 8 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. What is the minimum amount of vitamin A?

• A. 150
• B. 200
• C. 100
• D. 175

Answer 1

The correct option is A 150

Let x and y be the number of packets of food P and Q respectively. Obviously
x ≥ 0, y ≥ 0. The mathematical formulation of the given problem is as follows:
Minimise Z = 6x+3y
Constraint on calcium
12x + 3y ≥ 240, i.e. 4x + y ≥ 80 ... (1)
Constraint on iron
4x + 20y ≥ 460, i.e. x + 5y ≥ 115 ... (2)
Constraint on cholesterol
8x + 4y ≤ 300, i.e. 2x + y ≤ 75 ... (3)
x ≥ 0, y ≥ 0
Let’s graph the inequalities (1) to (4).
The feasible region (shaded) determined by the constraints (1) to (4) is shown in figure and note that it is bounded.
The coordinates of the corner points are $\left(\frac{5}{2},70\right)$, (15, 20) and $\left(\frac{260}{9},\frac{155}{9}\right)$
respectively. Let’s evaluate Z at these points

Corner Point	Z = 6x+3y
($\frac{5}{2}$, 70)	225
(15, 20)	150
($\frac{260}{9},\frac{155}{9}$)	$225$

From the table, Z is minimum at the point (15, 20). Hence, if 15 packets of food P and 20 packets of food Q are used in the special diet for minimising the value f vitamin A. The minimum amount of vitamin A will be 150 units.