Your company can make x-hundred grade A tires and y-hundred grade B tires a day, where 0≤x≤4 and y= (40-10x)/(5-x). The profit on each grade A tire is twice the profit on each grade B tire. Assuming all tires sell, what are the most profitable numbers of tire to make?

Answer 1

Profit = 2(x) + 1(40-10x)/(5-x)
= 2x + (40-10x)/(5-x)
= (2x^2 - 40)/(x-5)
d(profit)/dx = [(x-5)(4x) - (2x^2 - 40)(1) ]/(x-5)^2
= 0 for a max of profit

4x^2 - 20x - 2x^2 + 40 = 0

x^2 - 10x + 20 = 0
x = 5 ± √5
= appr 7.24 or appr 2.76
but x must be between 0 and 4, so x = 2.76
where x is in hundreds

so for a max profit they should make 276 A tires
and 552 B tires

check my arithmetic,

Answer 2

(3,0),(-2,6),(1,5),(3,6)

Answer 3

A tyre company can produce x hundreds of grade A tyres and y hundreds of grade B tyres, where and . The profit obtained by selling grades A tyres is twice that of grade B tyres. Find the number of grade A and grade B tyres produced so that the profit is maximum.