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Thread: Maximum value of combined function

  1. #1
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    Maximum value of combined function

    Hi,

    I hope someone can help. I'm currently working on 10c:

    Maximum value of combined function-screen-shot-2017-06-26-3.38.01-pm.png

    I know that the equation for the function is (25+x)(20,000-750x). In terms of finding the point of maximum revenue, I'm finding a hard time doing so. When I plot this on a graph, I get very large values. The answer according to the textbook is $25.83.

    Can someone help identify with me where the maximum point is for this equation?

    Thanks,
    Olivia
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  2. #2
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    Re: Maximum value of combined function

    Depending on the class you're currently in, I would suggest the following two methods:

    1) Convert the function to be in vertex form. The maximum value for a parabola that has a < 0 is at the vertex. If the vertex is (p,q) the maximum value will be q.
    2) Take the derivative of the function to find the x-value the maximum value occurs, then evaluate f(x)

    Let me know if you have further questions.
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  3. #3
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    Re: Maximum value of combined function

    you're looking at the function's maximum y-value ... the question is looking for the x-value where that occurs added to 25.

    $y = (25+x)(20,000-750x) = 500000+1250x - 750x^2$

    maximum occurs where $x = \dfrac{-b}{2a} = \dfrac{-1250}{-1500} = \dfrac{5}{6}$

    ticket price to maximize = $25 + \dfrac{5}{6} = \$25.83$
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  4. #4
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    Re: Maximum value of combined function

    Let's multiply out:

    $R = 500,000+1250x-750x^2 = -750\left( x^2-\dfrac{5}{3}x-\dfrac{2000}{3} \right)$

    R is revenue.

    Now, let's complete the square:

    $R = -750\left( x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}-\dfrac{2000}{3} \right) = -750\left(x-\dfrac{5}{6}\right)^2+\dfrac{3003125}{6}$

    This has a maximum at $x=\dfrac{5}{6}$ (because it is a parabola).

    So, it is maximized when the price is $25+x = 25+\dfrac{5}{6} \approx \$25.83$
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  5. #5
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    Re: Maximum value of combined function

    The formula that skeeter used, "x= -b/2a" can be derived by "completing the square", as SlipEternal did, in the general quadratic equation. ax^2+ bx+ c= a(x^2+ (b/a)x)+ c. [tex]a(x^2+ (b/a)x+ b^2/4a^2- b^2/4a^2)+ c= a(x^2+ (b/a)x+ b^2/4a^2)- b^2/4a+ c= a(x+ b/2a)^2+ (c- b^2/4a).

    If x= -b/2a then the quantity squared is 0 and the value of the function is c- b^2/4a.

    If x is any other number then, because a square is never negative, if a is negative, this is c- b^2/4a minus something so less than c- b^2/4a and c- b^2/4a is the maximum of the function. If a is positive, this is c- b^2/4a plus something so larger than c- b^2/4a and c- b^2/4a is the minimum value of the function.
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  6. #6
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    Re: Maximum value of combined function

    Great! This makes sense. Thanks for everyone's help.
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