Question

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.

Machine	Time (in minutes)
Type	I	II	III
A	12	18	6
B	6	0	9

Answer 1

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 \(=\frac{75}{100}x+\frac{50}{100}y\)\(\Rightarrow Z=\frac{3}{4}x+\frac{1}{2}y\)

On plotting the constraints, we get,

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

Hence, Proved.