Company's profit , derivative function

1. Oct 30, 2009

CanaBra

1. The problem statement, all variables and given/known data
A company's profit , in dollars, as a function of the number of items sold, x, is described by:

2. Relevant equations

15500(1-e^-0.0004x)
Calculate the profit when x=2500 units

3. The attempt at a solution

I am having a lots of problems with this one, but here is what I've attempted so far.
f(x)=15500d/dx(1-e^-0.0004x)
15500 (-0.0004x)(1-e)
-6.2xe^-0.0004
I am not sure if this is right or wrong, it is very confusing to me
Now, how am I supposed to find the value of x @ 2500?

2. Oct 30, 2009

Gib Z

You are thinking too hard! The function 15500(1-exp(-0.0004x)) is the company's profit in dollars, and you are told to find the profit when x=2500.

All you have to do is substitute x=2500 into the original function and evaluate!

3. Oct 30, 2009

CanaBra

Hi Gib ,

You've made my day, it is true, I can't believe it.
I will attempt it right away and post my answer.

Thank you so much

4. Oct 30, 2009

CanaBra

Here is my new attempt: I would appreciate if someone can check my answer:
P(x) =15500(1-e^-0.0004x)
evaluate at x=2500

15500(1-e^-0.0004(2500))
15500(1-e^-1)
e^-1=0.3678
P(2500) = 15500(1-0.3678)
15500-5700.9
P(2500) = $9,800 b) 12,000=15500(1-e^-0.0004x) 12,000/15500 = 1-e^-0.0004x e^-0.0004x=1-0.774193 e^-0.0004x=0.2258 e^-1.49=0.2258 e^-0.0004x=e^-1.49 -0.0004x=-1.49 x=-1.49/-0.0004 x=3,725 items 5. Oct 31, 2009 Gib Z How close do you want you answers? The method is correct except you could give it to more significant figures. For a), its$9797.87, and for b) its 3720 or 3721 depending on the question. It seems like its asking how many you would have to sell for $12,000 profit, so it would have to be 3721 items. 6. Oct 31, 2009 CanaBra Hi GIB Z, I didn' t pay attention to that, however looking at the question requirements it states that the answer needs to be to the rearest whole. Thus, I will use e^-1 = .3678794412, to get to P(2500) =$9798

NOW, for part (b), how do you get the amount of e^????
I looked at a table that I found online, but could not approximate as you did?
Is there a formula for this ?

THank you

7. Nov 1, 2009

Gib Z

Indeed there is!

If $$e^x = A$$ then $$x= \ln A$$ where "ln" is the Natural Logarithm, or
log base e. Its next to your "log" button on the calculator.