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Math Help - Please help!

  1. #1
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    Please help!

    Suppose that the demand x(in units) for a product is x=10,000-100p, where pd dollars is the market price per unit. The consumer expenditure for the product is E=px=10,000p-100p^2

    For what market price will expenditure be greatest?
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Please help!

    Are you familiar with finding the min or max of a function by setting its derivative equal to zero? Here you have E = 10000p-100p^2, so dE/dp = 10000-500p. Set this equal to zero, solve for p, and check that the result is a max and not a min by seeing if d^2E/dp^2 is positive or negative.

    If you are not famiiar with calculus you can try plotting the graph for E=10000p-100p^2 and see if you can estimate what value of p makes E the largest.
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  3. #3
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    Re: Please help!

    This is a quadratic function that does not require the "derivative" (or even knowing what a "derivative" [b]is[b]) to find max or min. Instead "complete the square". A "perfect square" is of the form (x- a)^2= x^2- 2ax+ a^2. Compare that to 10000p- 100p^2= -100(p^2- 100p). Since "p" corresponds to "a", "-100p" corresponds to -2ax, thus a corresponds to 50, and then a^2= 2500. To make p^2- 100p a perfect square we need to add 2500. And, of course, to keep the actual value the same, we need to subtract 2500 as well.

    That is, -100(p^2- 100p)= -100(p^2- 100p+ 2500- 2500)= -100(p^2- 100p+ 2500)+ 250000= -100(p- 10)^2+ 25000.

    Recalling that a square is never negative, it follows that -100(p- 10)^2 is never positive. So what is the maximum value for consumer expenditure and for what p does it have that maximum?
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Please help!

    Interesting technique - thanks for that. But a couple of typos crept into the equation. I've taken the liberty of making the corrections below in red:

    "That is,
    -100(p^2- 100p) .... = -100(p- {\color{red}5}0)^2+ 25000{\color{red}0}.

    Recalling that a square is never negative, it follows that -100(p- {\color{red}5}0)^2 is never positive.... "
    Last edited by ebaines; July 29th 2013 at 01:28 PM.
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