Question

A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of 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 80 on each piece of type A and 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? Make it as an LPP and solve graphically. What is the maximum profit per week?

Solution

Let x be the number of pieces manufactured of type A and y be the number of pieces manufactured of type B. Let us summarize the data given in the problem as follows:

Product	Time for Fabricating (in hours)	Time for Finishing (in hours)	Maximum labour hours available
Type A	9	1	180
Type B	12	3	30
Maximum Profit (in Rupees)	80	120

Thus, the mathematical form of above LPP is
Maximize Z = 80x+120y
subject to

9 straight x plus 12 straight y less or equal than 180 straight x plus 3 straight y less or equal than 30

Also, we have

straight x greater or equal than 0 comma space space straight y greater or equal than 0

Let us now find the feasible region, which is the set of all points whose coordinates satisfy all constraints.
Consider the following figure.

Thus, the feasible region consists of the points A, B and C.
The values of the objective function at the corner points are given below in the following table:

Points	Value of Z
A(12, 6)	Z = 80 x 12 + 120 x 6 = Rs. 1680
B(0, 10)	Z = 80 x 0 +120 x 10 = Rs. 1200
C(20, 0)	Z = 80 x 20 + 120 x 0 = Rs.1600

Clearly,Z is maximum at x=12 and y=6 and the maximum profit is Rs.1680.

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Three Dimensional Geometry

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