Question

A chemical company produces a chemical containing three basic elements A, B, C, so that it has at least 16 litres of A, 24 litres of B and 18 litres of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 litres of A, 12 litres of B and 2 litres of C. Each unit of compound II has 2 litres of A, 2 litres of B and 6 litres of C. The cost per unit of compound I is ₹ 800 and that of compound II is ₹ 640. Formulate the problems as LPP and solve it to minimize the cost.

Answer 1

Let the company buy x units of compound I and y units of compound II.

Then the total cost is z = ₹ (800x + 640y)

This is the objective function that is to be minimized.

The constraints are as per the following table:

	Compound I (x)	Compound II (y)	Minimum requirement
Element A	4	2	16
Element B	12	2	24
Element C	2	6	18

From the table, the constraints are 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18.

Also, the number of units of compound I and compound II cannot be negative.

∴ x ≥ 0, y ≥ 0

∴ The mathematical formulation of given LPP is

Minimize z = 800x + 640y, subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x ≥ 0, y ≥ 0.

First we draw the lines AB, CD and EF whose equations are 4x + 2y = 16, 12x + 2y = 24 and 2x + 6y = 18 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	4x + 2y = 16	A(4, 0)	B(0, 8)	≥	non-origin side of line AB
CD	12x + 2y = 24	C(2, 0)	D(0, 12)	≥	non-origin side of line CD
EF	2x + 6y = 18	E(9, 0)	F(0, 3)	≥	non-origin side of line EF

The feasible region is shaded in the graph.

The vertices of the feasible region are E(9, 0), P, Q and D(0, 12).

P is the point of intersection of the lines

2x + 6y = 18       ...(1)

and 4x + 2y = 16      ...(2)

Multiplying equation (1) by 2, we get

4x + 12y = 36

Subtracting equation (2) from this equation, we get

10y = 20

∴ y = 2

∴ From (1), 2x + 6(2) = 18

∴ 2x = 6

∴ x = 3

∴ P = (3, 2)

Q is the point of intersection of the lines

12x + 2y = 24      ...(3)

and 4x + 2y = 16      ...(2)

On subtracting, we get

8x = 8     

∴ x = 1

∴ From (2), 4(1) + 2y = 16

∴ 2y = 12

∴ y = 6

∴ Q = (1, 6)

The values of the objective function z = 800x + 640y at these vertices are

z(E) = 800(9) + 640(0) = 7200 + 0 = 7200

z(P) = 800(3) + 640(2) = 2400 + 1280 = 3680

z(Q) = 800(1) + 640(6) = 800 + 3840 = 4640

z(D) = 800(0) + 640(12) = 0 + 7680 = 7680

∴ The minimum value of z is 3680 at the point (3, 2).

Hence, the company should buy 3 units of compound I and 2 units of compound II to have the minimum cost of ₹ 3680.