Question

A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20 per unit of A and ₹ 30 per unit of B. Both A and B make use of two essential components a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each units of B requires 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should be manufactured per month to maximize profit? How much is the maximum profit?

Answer 1

Let the firm manufactures x units of item A and y units of item B.

Firm can make profit of ₹ 20 per unit of A and ₹ 30 per unit of B.

Hence, the total profit is z = ₹ (20x + 30y)

This is the objective function which is to be maximized. The constraints are as per the following table:

	Item A (x)	Item A (y)	Total supply
Motor	3	2	210
Transformer	2	4	300

From the table, the constraints are

3x + 2y ≤ 210, 2x + 4y ≤ 300

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

Hence, the mathematical formulation of given LPP is :

Maximize z = 20x + 30y, subject to

3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0

We draw the lines AB and CD whose equations are

3x + 2y = 210 and 2x + 4y = 300 respectively.

Line	Equation	Points on the X-axis	Points on the Y-axis	Sign	Region
AB	3x + 2y = 210	A(70, 0)	B(0, 105)	≤	origin side of line AB
CD	2x + 4y = 300	C(150, 0)	D(0, 75)	≤	origin side of line CD

The feasible region is OAPDO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), A (70, 0), P and D (0, 75).
P is the point of intersection of the lines

2x + 4y = 300      ....(1)

and 3x + 2y = 210       ....(2)

Multiplying equation (2) by 2, we get

6x + 4y = 420 

Subtracting equation (1) from this equation, we get

∴ 4x = 120 

∴ x = 30

Substituting x = 30 in (1), we get

2(30) + 4y = 300

∴ 4y = 240

∴ y = 60

∴ P is (30, 60)

The values of the objective function z = 20x + 30y at these vertices are

z(O) = 20(0) + 30(0) = 0 + 0 = 0

z(A) = 20(70) + 30(0) = 1400 + 0 = 1400

z(P) = 20(30) + 30(60) = 600 + 1800 = 2400

z(D) = 20(0) + 30(75) = 0 + 2250 = 2250

∴ z has the maximum value 2400 when x = 30 and y = 60

Hence, the firm should manufactured 30 units of item A and 60 units of item B to get the maximum profit of  ₹ 2400 frim