Posted by ibranian on Tuesday, August 7, 2012 at 5:16am.
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, Tuesday, August 7, 2012 at 8:22am
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, Tuesday, August 14, 2012 at 3:17pm
(3,0),(2,6),(1,5),(3,6)
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