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Math Help - More optimization...

  1. #1
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    More optimization...

    Given C(x) = 680 + 4x + 0.01x^2 and demand is p(x) = 12 - x/500, find the value of x that will maximize profit.

    I assume that I somehow need to find the minimum for C(x) and the max for p(x), but I'm not sure if I somehow have to combine the two or something, or if the x has to be the same for both functions...
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  2. #2
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    If you think about it, profit is Revenue minus Cost. Also, Revenue must be Demand times Price. Mysteriously, you don't have a price, but your revenue is p instead of the more usual q. Hoping that p really is your demand and not a price you haven't included, I'm going to call it q, and call the price A. Then,

    P\ =\ R - C\ =\ Aq - C

    Just in case a picture helps...



    ... where straight continuous lines differentiate downwards (integrate up) with respect to x.

    Spoiler:

    Put dP/dx = 0 and then make x the subject in the bottom row.

    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; August 3rd 2010 at 01:15 PM. Reason: Hoping
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  3. #3
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    So x = -50A/500 - 200?
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  4. #4
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    That was my conclusion, but could you post the whole problem? I think demand should depend on price, not quantity (i.e. demand is quantity, hence q), so something's missing after all.
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  5. #5
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    I did post the whole problem. That was all there was. "Given C(x) = 680 + 4x + 0.01x^2 and demand is p(x) = 12 - x/500, find the value of x that will maximize profit."

    With just the cost function and the demand function given I'm not sure how to turn those into the profit function and then get the answer.
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  6. #6
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    I suppose our result is ok as long as x isn't quantity sold. It could be, say, number of pages in a book, which might affect the cost and the demand.
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  7. #7
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    True. It gives no context so I assume that's what it meant...
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  8. #8
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    On the other hand, if the max profit is going to be for a value of x other than zero, then x must go less than zero, so pages wouldn't work!
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  9. #9
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    There has to be something we're missing...I mean, it has to have an actual value right?
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  10. #10
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    Okay, so I confirmed that the answer we came to isn't correct. Does anyone know what we did wrong?
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  11. #11
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    Okay, so a minimum value for the C(x) function is -200, which would be the value of x that minimizes cost the most, right? Well, I can't find it for p(x), on top of the fact that that doesn't answer the question of how to maximize the profit without a profit function....help?
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  12. #12
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    Bump...-200 isn't correct either.
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