Posted by **ibranian** on Tuesday, August 7, 2012 at 5:16am.

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?

- math -
**Reiny**, Tuesday, August 7, 2012 at 8:22am
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,

- math -
**lamar henderson**, Tuesday, August 14, 2012 at 3:17pm
(3,0),(-2,6),(1,5),(3,6)

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