# Need help with some question

• November 3rd 2009, 06:26 PM
darksnakex
Need help with some question
A company has found that if it produces x thousand units/day of a product, The total cost would be T(x) thousand \$, where T(x) = x^3 - 6x^3 + 13x + 15, when 0<x<9. They sell it for 28 \$/unit.

a) State the profit V(x) thousand \$ as a function of x.
b) Find the maximum profit per day.
• November 3rd 2009, 07:00 PM
Keithfert488
I'm assuming you meant to type $T(x)=x^3-6x^2+13x+15$
So profit is sales-cost. Therefore, $V(x)=28000x-(x^3-6x^2+13x+15)=-x^3+6x^2+27987x-15$

To find the maximum, you must first differentiate giving you $V'(x)=-3x^2+12x+27987$.
Then find the critical numbers by setting V'(x) to 0.

$0=x^2-4x-9329$
Quadratic formula gives you the critical values of x=98 and x=-94 but those aren't in the interval (0,9). So the maximum is when x is either 0 or 9. You can eliminate 0 because they would obviously make no profit if they didn't make a product. So plug x=9 into the profit equation.

$V(9)=-9^3+6*9^2+27987*9-15=-729+486+251883-15=251625$
The maximum profit is \$251625
• November 3rd 2009, 07:11 PM
darksnakex
The answer to a) was -x^3 -6x^2 +15x -15. Thats because they used 28x, instead of 28000x.

After that i couldn't get the answer to b), they get 85 000. I dont really understand how.
• November 3rd 2009, 07:24 PM
Keithfert488
So $V(x)=-x^3-6x^2+15x-15$

First you need to find the critical numbers and see if they are in your interval (0,9). To find the critical numbers, derive the equation and set it equal to 0.

$V'(x)=-3x^2-12x+15$
$0=x^2+4x-5=(x+5)(x-1)$
$x=1$ $x=-5$

So you have a critical value at x=1. To find the maximum, you must plug in the endpoints of your interval (i.e. x=0 and x=9) and the critical number within the interval (i.e. x=1) into the original equation and find which one is the highest. That is your maximum.

$V(0)=-15$
$V(1)=-1-6+15-15=-7$
$V(9)=-9^3+6*9^2+15*9-15=-729+486+135-15=-123$

Well, that's what I got and it doesn't make sense. Can anyone else help? I am pretty sure that I have the right method of getting the maximum. I just don't understand how the answer could be negative.
• November 3rd 2009, 07:49 PM
darksnakex
I have figured out where i was wrong! Thanks anyways!
• November 3rd 2009, 09:25 PM
fandong12345
Thank you very much!
Thank you very much!