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Math Help - Net profit Expectation Question

  1. #1
    Junior Member
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    Net profit Expectation Question

    Hi all,

    I am having trouble with the problem:

    A manufacturer ships car radios in cartons of n radios. The profit per radio is 59.50 per radio, and it costs 25 dollars to ship a carton. To promote sales and quality assurance, the retailed promises to pay the buyer 200X^2 for X radios that are defective. Suppose radios are produced independently and that 5% of radios are defective. How many radios should be placed per carton to maximize net profit.


    So I have that the net profit described by (59.50*n) - 25 - 200 (0.05n)^2
    where n is the number of boxes in a carton.

    If I compute the expectation of this I get f representing the net profit:

    f(x)= 59.50 * n -25 - 200 E(X^2) = 59.50n-25-200 (0.05n)^2
    (since X follows a binomial distribution with p = 0.05)

    Computing the derivative of this function ( to find maximum point) I get n = 59.50. However, the answer is 50. I am not sure what I failed to consider in this question.
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  2. #2
    MHF Contributor
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    Re: Net profit Expectation Question

    59.50n-25-200 (0.05n)^2
    The stuff in red is \left(E[X] \right)^2, it should be E[X^2]

    Correct that and see if you end up with the right answer. If you dont know E[X^2] you can calculate it using: E[X^2] = Var(X) + \left(E[X] \right)^2
    Last edited by SpringFan25; November 9th 2011 at 03:30 AM.
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