Question

A firm manufactures two products A and B on which profit earned per unit Rs 3/- and Rs 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 minute 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 product A and B to be manufactured to get the maximum profit.

Answer 1

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

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

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

This is the linear function which is to be maximized. 

Hence, it is the objective function. 

The constraints are as per the following table :

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.

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 1800 when x = 0 and y = 450 Hence, the firm gets maximum profit of Rs 1800 if it manufactures 450 units of product B and no unit product A.