Question

A chemist has a compound to be made using 3 basic elements X, Y, Z so that it has at least 10 litres of X, 12 litres of Y and 20 litres of Z. He makes this compound by mixing two compounds (I) and (II). Each unit compound (I) had 4 litres of X, 3 litres of Y. Each unit compound (II) had 1 litre of X, 2 litres of Y and 4 litres of Z. The unit costs of compounds (I) and (II) are ₹ 400 and ₹ 600 respectively. Find the number of units of each compound to be produced so as to minimize the cost

Answer 1

Let the chemist produce x units of compound I and y units of compound II.

Since x and y cannot be negative, x ≥ 0, y ≥ 0

The unit costs of compounds I and II are ₹ 400 and ₹ 600 respectively.

Total Cost = Z = 400x + 600y

We construct a table with constraints of X, Y and Z as follows:

	Compound I	Compound II	Least value
X	4	1	10
Y	3	2	12
Z	–	4	20

From the table, the constraints are

4x + y ≥ 10

3x + 2y ≥ 12

4y ≥ 20

∴ Given problem can be formulated as follows:

Minimize Z = 400x + 600y

Subject to 4x + y ≥ 10

3x + 2y ≥ 12

4y ≥ 20, x ≥ 0, y ≥ 0

To draw the feasible region, construct table as follows:

Inequality	4x + y ≥ 10	3x + 2y ≥ 12	4y ≥ 20
Corresponding equation (of line)	4x + y = 10	3x + 2y = 12	4y = 20
Intersection of line with X-axis	`(5/2, 0)`	(4, 0)	–
Intersection of line with Y-axis	(0, 10)	(0, 6)	(0, 5)
Region	Non-Origin side	Non-Origin side	Non-Origin side

Shaded portion EABY is the feasible region, whose vertices are A and B(0, 10).

A is the point of intersection of the lines 4y = 20 and 4x + y = 10,

Solving the above equations, we get

x = `5/4`, y = 5

∴ A ≡ `(5/4, 5)`

Here, the objective function is

Z = 400x + 600y

∴ Z at A`(5/4, 5) = 400(5/4) + 600(5)`

= 500 + 3000

= 3500

Z at B (0, 10) = 400(0) + 600(10)

= 6000

∴ Z has minimum value 3500 at x = `5/4` and y = 5.

∴ The chemist should produce `5/4` units of compound I and 5 units of compound II to minimize the cost.