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Company's profit , derivative function

  1. Oct 30, 2009 #1
    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?
    Please help!!
     
  2. jcsd
  3. Oct 30, 2009 #2

    Gib Z

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    Homework Helper

    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!
     
  4. Oct 30, 2009 #3
    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
     
  5. Oct 30, 2009 #4
    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
     
  6. Oct 31, 2009 #5

    Gib Z

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    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.
     
  7. Oct 31, 2009 #6
    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
     
  8. Nov 1, 2009 #7

    Gib Z

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    Indeed there is!

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