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Question

A manufacturing company makes two types of teaching aids A and B of Mathematics for Class X. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs. 80 on each piece of type A and Rs. 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch.

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Solution

Let the pieces of type A manufactured per week be x.

Let the pieces of type B manufactured per week be y.

maximum profit z=80x+120y

Fabricating hours for A is 9 and finishing hours is 1

Fabricating hours for B is 12 and finishing hours is 3

Maximum number of fabricating hours is 180.

9x+12y180

3x+4y60

Maximum number of finishing hours is 30.

x+3y30

Thus we can formulate the Linear Programming Problem as follows:

maxz=80x+120y subjected to the constraints, 3x+4y60,x+3y30,x0,y0

Solving the constraints we get x=12,y=6

Now the area of feasible region is as shown in the figure.


We can see that the points bounded by the feasible region are A(0,10),B(12,6),C(20,0)

Now let us calculate the maximum profit.

At A(0,10),z=80(0)+120(10)=1200

At B(12,6),z=80(12)+120(6)=1680

At C(20,0),z=80(20)+120(0)=1600

Hence the maximum profit is at (12,6)

Therefore 12 pieces of type A and 6 pieces of type B should be manufactured per week to get a maximum profit of Rs.1680

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