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Math Help - Quadratic equation problem/optimization

  1. #1
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    Quadratic equation problem/optimization

    "Sweet Harmony crafts has determined that when x hundred dulcimers are built, the average cost per dulcimer can be estimated by:
    C(x) = 0.1x^2 - 0.7x + 2.425,
    where C(x) is in hundreds of dollars. What is the minimum average cost per dulcimer and how many dulcimers should be built in order to achieve that minimum?"

    Just need some guidance on how to interpret this; what I should be doing to find out the answers to the questions in bold. Thanks!
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  2. #2
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    Differentiate the equation to get C'(x).

    Then find the point where C'(x) = 0 and this will be the minimum average cost.
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  3. #3
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    Quote Originally Posted by Educated View Post
    Differentiate the equation to get C'(x).

    Then find the point where C'(x) = 0 and this will be the minimum average cost.
    Sorry man, could you explain to me what "differentiate the equation" means? I have never heard that term come up in my class or paper.

    As for C(x) = 0, is that the same concept of finding the x-intercepts? (I.E.: Set 0.1x^2 - 0.7x + 2.425 equal to 0, then use the Quadratic Formula)
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  4. #4
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    Quote Originally Posted by Kyrie View Post
    Sorry man, could you explain to me what "differentiate the equation" means? I have never heard that term come up in my class or paper.

    As for C(x) = 0, is that the same concept of finding the x-intercepts? (I.E.: Set 0.1x^2 - 0.7x + 2.425 equal to 0, then use the Quadratic Formula)
    Differentiation requires some calculus. Seeing you haven't done that, for a quadratic equation of the form

     a^2x + bx + c = 0

    If a > 0, the minimum point is when x = -b/(2a)
    If a < 0, -b/(2a) gives the x intercept of the maximum point.
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  5. #5
    Senior Member Educated's Avatar
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    Quote Originally Posted by Kyrie View Post
    As for C(x) = 0, is that the same concept of finding the x-intercepts? (I.E.: Set 0.1x^2 - 0.7x + 2.425 equal to 0, then use the Quadratic Formula)
    No. C'(x) isn't the same as C(x). C'(x) is the derivative of the function C(x), where C'(x) is the formula for the change in y over the change in x, which is the gradient/slope of C(x) at a given point.

    Since you haven't learnt this yet, use Gusbob's method to find the minimum point.
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