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Math Help - Real-life application of Calculus

  1. #1
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    Exclamation Real-life application of Calculus

    Hello all,


    I am currently doing some real-life applications for calculus and stumbled upon the following question, can anyone please help me?



    When a chemical factory produces a certain compound they have observed that the daily amount y of defective compounds depends on the total amount x that they produce according to the empirical formula:
    y = 0.02x + 0.00005x^2
    where x and y are measured in kilograms. The manufacturer earns $100 per kilogram of nondefective compound and incurs a loss of $20 per kilogram of defective compound.
    (a) Write down a function for the total profit P(x).
    (b) How many kilograms should they produce to maximise their profits?


    Any help is much appreciated,


    Dranalion
    Last edited by Dranalion; October 2nd 2009 at 05:51 AM.
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  2. #2
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    I hope I understand your question correctly.
    for every xkg of this compoud, ykg is defective. so they lose $20y for every x amount they produce and earn $100(x-y). so their profit is $100(x-y)-$20y. the next find the maximum point of the graph of P(x) by finding the stationary points
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  3. #3
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    How do I find the function, P(x)? Does finding the stationary points allow me to do so?
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  4. #4
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    Quote Originally Posted by Dranalion View Post
    How do I find the function, P(x)? Does finding the stationary points allow me to do so?
    P(x) is the profit, that is, amount they earn minus what they lose. so P(x)=100(x-y)-20y
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  5. #5
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    How about Part B of the question:

    (b) How many kilograms should they produce to maximise their profits?

    Any help is much appreciated,

    Dranalion
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  6. #6
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    Sorry, substituting y = 0.02x + 0.00005x^2 into P(x), the result is:

    P(x) = 97.6x + 0.0006x^2

    P'(x) = 97.6 + 0.012x (for finding the stationary points, then allowing that to equal zero to find the maximum profits

    Is this correct? Am I on the right track?
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  7. #7
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    yes, you got it!
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