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Math Help - Forming an equation and differentiating it

  1. #1
    Senior Member Mukilab's Avatar
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    Forming an equation and differentiating it

    So you need to find an optimum price, so I just thought you had to find a stationary point, i.e. when the differentiation equals 0

    But if you differentiate Q you just get -80. So you can't set this to 0, nor can the optimum price be -80. So I thought, ok it must be a bit harder than that.

    So the cost per unit 'a' = 300,000 + 20a since there is a fixed cost of 300000 and a variable cost of 20 per unit a

    But I just don't know where to go from here since cost per unit is not necessarily the price per unit, so we can't sub it into the quantity demanded equation. Even if we did, it wouldn't matter because we would still get 0=a number when we differentiate. I think that you have to somehow set up a quadratic in terms of price per unit and then differentiate THAT. But I just don't know how to do it. Any help?

    Screenshot of the problem:
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  2. #2
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    Re: Forming an equation and differentiating it

    Multiply the profit per unit by the quantity demanded to get the total profit on the units sold. You can then differentiate the expression for total profit to find a maximum.
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  3. #3
    Senior Member Mukilab's Avatar
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    Re: Forming an equation and differentiating it

    But I don't know the profit per unit, only the cost to produce one

    Profit per unit would be P - 300000 + 20a i.e. Price minus cost.
    But then this would give us two variables if we multiply that by Quantity demanded
    Last edited by Mukilab; May 27th 2013 at 01:27 AM.
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  4. #4
    Senior Member Mukilab's Avatar
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    Re: Forming an equation and differentiating it

    What if I sub in quantity demanded per unit, into the formula I wrote for cost per unit.

    So 300,000 + 20a = 300,000 + 20(15000-80P)
    and then times profit per unit = P - 300,000 + 20(15,000 - 80P) by quantity demanded

    This would give me P = 187.6 rounded, after differentiation. Is this correct?

    But when I do differentiate again (to find the nature of this stationary point) it turns out it is a minimum, not a maximum so it must be wrong, given the question
    Last edited by Mukilab; May 27th 2013 at 01:38 AM.
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  5. #5
    Senior Member Mukilab's Avatar
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    Re: Forming an equation and differentiating it

    Nevermind I got it

    I got Profit per units produced = Pa - 300,000 - 20a where a is the number of units
    then subbed in quantity demanded for a to get total profit

    then I got P=103.75 by calculating and differentiating, which is a maximum
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