Q. 20

A manufacture makes two types, A and B, of teapots. Three machines are needed for the manufacture and the time required for each teapot on the machines is given below.

Each machine is available for a maximum of 6 hours per day. If the profit on each teapot of type A is 75 paise and that on each teapot of type B is 50 paise, show that 15 teapots of type A and 30 of type B should be manufactured in a day to get the maximum profit.


Answer :

Let x teapots of type A and y teapots of type B manufactured.


Then,


x ≥ 0, y ≥ 0


Also,


12x + 6y ≤ 6 × 60


12x + 6y ≤ 360


2x + y ≤ 60…..(1)


And,


18x + 0y ≤ 6 × 60


X ≤ 20……(2)


Also,


6x + 9y ≤ 6 × 60


2x + 3y ≤ 120…..(3)


The profit will be given by: Z


On plotting the constraints, we get,



Profit will be maximum when x = 30 and y = 15


Hence, Proved.


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