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in Linear Programming by (25.9k points)
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A firm manufacturing two types of electrical items A and B, can make a profit of Rs 20/- per unit of A and Rs 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 the manufacture per month to maximize profit? How much is the maximum profit?

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

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

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

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

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.

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.

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