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.
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.