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Math Help - Optimization of eponential functions

  1. #1
    Newbie darkfenix21's Avatar
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    Optimization of eponential functions

    Hi everyone just wondering if someone could give me a push in the right direction for a problem I'm working on. I've been given the statement:

    P(x)=10^6(1+(x-1)e^-0.001x)

    Where P is profit and x is the number of items being produced. It wants to know what the max profit is when the domain is [0,2000]. Now I've concluded to do so I must take the derivative of this.Yet I'm not exactly sure how to go about doing that. So if anyone could purhaps break that down into a step by step format for me it would be much appreciated. The answer is given at the back of the book as x=1001

    Thank you in advance

    After some more time looking at it I can see where my derivative went wrong. For those interested I will post my solution.

    10^6[1+(e^-0.001x)(x-1)] (rearranged)

    P'(x)= 10^6[(e^-0.001x) (1) + (x-1)(e^-0.001x)(-0.001)] (Use the product and chain rules)

    =10^6[(e^-0.001x) +(-0.001xe^-0.001x)(x-1)] (simplifying step)

    =10^6[(e^-0.001x) +(0.001e^-0.001)(-0.001xe^-0.001)] (we can now remove any common factors in this case e^-0.001)

    =(10^6)(e^-0.001)[1.001 - .001x] (run the 10^6 through the brackets)

    =(e^-0.001)[1.001x10^6 -1000x] (To find the critical points we set this =0 and since (e^-0.001) can never be 0 we ignore it)

    0=[1.001x10^6-1000x] (solve for x)

    1.001x10^6=1000x

    x=1001

    now plug 1001 into the original equation to get your answer. Also you must do the domain points of 0 and 2000 to ensure you have the correct answer

    so P(0)=0 P(1001)= 3.685x10^9 P(2000)=2.71x10^9


    Thank you to all who took the time to read and try an puzzle this out it is much appreciated
    Last edited by darkfenix21; April 19th 2010 at 09:47 AM. Reason: solved
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  2. #2
    Senior Member apcalculus's Avatar
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    If the function:
    P(x)=10^6(1+(x-1)e^{-0.001x})

    is typed correctly, then the derivative is equal to the derivative of:

    P(x)=10^6(x-1)e^{-0.001x}

    We need to use the product rule here:

    P'(x)=10^6[e^{-0.001x}+(x-1)e^{-0.001x}(-0.001)]

    Factor the derivative:

    P'(x)=10^6e^{-0.001x}[1+(x-1)(-0.001)]

    P'(x)=10^6e^{-0.001x}[2.001-0.001x]

    Solving for the critical number you get:

    x=2001.

    so clearly something went wrong in my rushed reply here, because this does not agree with your book's answer. or... the original equation could be different and I misinterpreted the parentheses. I hope the procedure helps though.

    Because you have a closed interval for the domain. You should compare the population values at the end points to the ones at the critical numbers.

    Good luck!
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