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Math Help - Marginal Profit Question

  1. #1
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    Marginal Profit Question

    Im having a lot of trouble trying to figure out how to solve this problem, any help is greatly appreciated! Thanks.

    The profit (measured in dollars) of a small company has been determined to be
    P(x)= 20000 (x / 100 + x^2), where x is the production level (in units).

    a) Compute the marginal profit

    b) Find the production level xmax that leads to the maximal profit within the range
    0 ≤ x ≤ 30.

    c) How many dollars of profit does the company make at that production level?

    d) Does your answer change if x has to be within the range 0 ≤ x ≤ 10 instead? If yes, how?
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  2. #2
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    a)
    the marginal profit is \frac{dP}{dx}

    b)
    maximal profit occurs either at the point where marginal profit = 0, or x=0, or x=30.

    Find the profit at each of these points and see which one is higher
    hintThere is no point in the range 0 < x < 30 where marginal profit =0 in this case.


    c)
    this is equal to P(x), where x is the production level you found in part b

    d)
    if your existing answer is in the range 0 < x< 10, then it will not change.

    However if your existing answer is not in the range, you need to find a new one, which will be either:
    x=0
    x=10
    or somewhere in between where the marginal profit is zero (you can discount this possibility as if there was a point in the range with marginal profit =0, you would have found it in part b)
    Last edited by SpringFan25; June 25th 2010 at 01:52 AM.
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  3. #3
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    Thanks so much for your help!

    Please let me know if these answers are correct:
    A.)20000 [(100-x^2)/(100+x^2)^2]
    B.)x=10
    c)p=1000
    d.) No, because in the previous problem, 10 had the highest amount of profit.
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  4. #4
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    hmm, if i understood your function correctly i dont think you have the marginal profit right

    You wrote P(x)= 20000 (x / 100 + x^2)

    which is P(x) = 20000 \left(0.01x + x^2 \right) = 200x + 20000x^2

    Did you mean

    P(x) = 20000 \left(\frac{x}{100 + x^2} \right) ?


    I'll assume you meant P(x) = 20000 \left(0.01x + x^2 \right) = 200x + 20000x^2
    a)
    P(x) = 20000 \left(0.01x + x^2 \right) = 200x + 20000x^2
    MP = \frac{dP}{dx} = 200 + 40000x


    b) MP = 0 has no solutions where x > 0
    So max pforit occurs at either x=0 or x=30
    p(0) = 0
    p(30) = 1200200

    so x=30 maximises profit

    c)
    p(30) = 1200200

    d)
    Yes, the new maximum is at x=10
    Last edited by SpringFan25; June 25th 2010 at 02:08 PM.
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