Posted by ibranian on .
Your company can make xhundred grade A tires and yhundred grade B tires a day, where 0≤x≤4 and y= (4010x)/(5x). 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?

math 
Reiny,
Profit = 2(x) + 1(4010x)/(5x)
= 2x + (4010x)/(5x)
= (2x^2  40)/(x5)
d(profit)/dx = [(x5)(4x)  (2x^2  40)(1) ]/(x5)^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, 
math 
lamar henderson,
(3,0),(2,6),(1,5),(3,6)