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 minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?

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Solution

Let x units of food P and y units of food Q be mixed in the diet.
Quantity of food cannot be negative.
$∴ x, y \geq 0$

The given information can be tabulated as follows:

Calcium Iron Cholestrol Vitamin A

P 12 4 6 6

Q 3 20 4 3

Requirement 240 460 300 is to be minimised

The mathematical formulation of the given LPP is

Min Z = $6 x + 3 y$

subject to

$12 x + 3 y \geq 240 4 x + 20 y \geq 460 6 x + 4 y \leq 300$

First we will convert inequations into equations as follows:
12x + 3y = 240, 4x + 20y = 460, 6x + 4y = 300, x = 0 and y = 0

Region represented by 12x + 3y ≥ 240:
The line 12x + 3y = 240 meets the coordinate axes at $A (20, 0)$ and B(0, 80) respectively. By joining these points we obtain the line
12x + 3y = 240. Clearly (0,0) does not satisfies the 12x + 3y = 240. So, the region which does not contain the origin represents the solution set of the inequation 12x + 3y ≥ 240.

Region represented by 4x + 20y ≥ 460:
The line 4x + 20y = 460 meets the coordinate axes at C(115, 0) and $D (0, 23)$ respectively. By joining these points we obtain the line
4x + 20y = 460. Clearly (0,0) does not satisfies the inequation 4x + 20y ≥ 460. So,the region which does not contain the origin represents the solution set of the inequation 4x + 20y ≥ 460.

Region represented by 6x + 4y ≤ 300:
The line 6x + 4y = 300 meets the coordinate axes at E(50, 0) and $F (0, 75)$ respectively. By joining these points we obtain the line
6x + 4y = 300. Clearly (0,0) satisfies the inequation 6x + 4y ≤ 300. So,the region which contains the origin represents the solution set of the inequation
6x + 4y ≤ 300.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints 12x + 3y ≥ 240, 4x + 20y ≥ 460, 6x + 4y ≤ 300, x ≥ 0 and y ≥ 0 are as follows:

The corner points are G(2, 72), H(40, 15), I(15, 20)

The value of the objective function at the corner points

Corner points Z = 6x + 3y

G(2, 72) 228

H(40, 15) 285

I(15, 20) 150

The minimum value of Z is 150 which is attained at I $(15, 20)$ .

Thus, for minimum vitamin A i.e. 150 units , 15 packets of food P and 20 packets of food Q should be used.

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