Question

A firm manufactures two products A and B on which profit earned per unit are ₹ 3 and ₹ 4 respectively. Each product is processed on two machines M₁ and M₂. The product A requires one minute of processing time on M₁ and two minutes of processing time on M₂, B requires one minute of processing time on M₁ and one minute of processing time on M₂. Machine M₁ is available for use for 450 minutes while M₂ is available for 600 minutes during any working day. Find the number of units of products A and B to be manufactured to get the maximum profit.

Answer 1

Let the firm manufacture x units of product A and y units of product B.

The profit earned per unit of A is ₹ 3 and B is ₹ 4.

Hence, the total profit is z = ₹ (3x + 4y)

This is the linear function that is to be maximized.

Hence, it is an objective function.

The constraints are as per the following table:

Machine	Product A (x)	Product B  (y)	Total availability of time (minutes)
M1	1	1	450
M2	2	1	600

From the table, the constraints are x + y ≤ 450, 2x + y ≤ 600.

Since, the number of gift items cannot be negative, x ≥ 0, y ≥ 0.

∴ The mathematical formulation of LPP is,

Maximize z = 3x + 4y, subject to x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0.

Now, we draw the lines AB and CD, whose equations are x + y = 450 and 2x + y = 600, respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	x + y = 450	A(450, 0)	B(0, 450)	≤	origin side of line AB
CD	2x + y = 600	C(300, 0)	D(0, 600)	≤	origin side of line CD

The feasible region is OCPBO which is shaded in the graph.

The vertices of the feasible region are O(0, 0), C(300, 0), P and B(0, 450).

P is the point of intersection of the lines

2x + y = 600          ...(1)

and x + y = 450     ...(2)

On subtracting, we get

∴ x = 150 

Substituting x = 150 in equation (2), we get

150 + y = 450

∴ y = 300

∴ P ≡ (150, 300)

The values of the objective function z = 3x + 4y at these vertices are

z(O) = 3(0) + 4(0) = 0 + 0 = 0

z(C) = 3(300) + 4(0) = 900 + 0 = 900

z(P) = 3(150) + 4(300) = 450 + 1200 = 1650

z(B) = 3(0) + 4(450) = 0 + 1800 = 1800

∴ z has the maximum value of 1800 when x = 0 and y = 450.

Hence, the firm gets a maximum profit of ₹ 1800 if it manufactures 450 units of product B and no units of product A.