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Math Help - profit, percentage, differentiation

  1. #1
    Senior Member furor celtica's Avatar
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    profit, percentage, differentiation

    The manager of a supermarket usually adds a mark-up of 20% to the wholesale prices of all the goods he sells. He reckons that he has a loyal core of F customers and that, if he lowers his mark-up to x% he will attract an extra k(20 x) customers from his rivals. Each week the average shopper buys goods whose wholesale value is A. Show that with a mark-up of x% the supermarket will have an anticipated weekly profit of
    Ax/100 ((F+20k) kx)
    The only thing I dont understand here is the 1/100. I get that: total profit= profit per customer x number of customers = (A + Ax A) (F + k(20-x)) = Ax (F + 20k kx).
    But I dont get where the 1/100 comes in, I know its cos its a percentage but I dont see where to insert it, mathematically.

    Show that the manager can increase his profit by reducing his mark-up below 20% provided that 20k>F.

    Here I differentiated weekly profit = f(x) with f(x) = 1/100 (AFx + 20Akx AK(x)^2) to get
    f(x) = 1/100 (AF + 20Ak 2Akx) > 0
    AF + 20Ak 2Akx > 0
    We know A > 0 so we can divide both sides by A; F + 20k > 2kx
    We know k > 0 so we can divide both side by 2k; x< (F + 20k)/2k
    x> -(F + 20k)/2k
    x must be smaller than 20 (given, mark-up below 20%)
    -(F+20k)2k < 20
    40k > -F -20k
    -F< 60k
    ?
    Can someone show me where I went wrong?
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  2. #2
    Senior Member furor celtica's Avatar
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    ok here's some more work
    Last edited by furor celtica; August 23rd 2010 at 05:07 AM.
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  3. #3
    Senior Member furor celtica's Avatar
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    ok i figured out the percentage part, i.e. the first part of the question.
    in the second part i can prove the statement if i can prove that F+20k < 40k
    i already have F=20k> 2kx and x<20 so i have to prove 2kx<(F+20k)<40k
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