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anonymous

  • 5 years ago

how to find the minimum value of this quadratic equation?

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  1. heisenberg
    • 5 years ago
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    which quadratic function would that be?

  2. anonymous
    • 5 years ago
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    C(x)=.045x^2-110x+10000

  3. heisenberg
    • 5 years ago
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    this seems like you'll want to use calculus. finding the max/min values using the first derivative test. do you know derivatives/calculus?

  4. anonymous
    • 5 years ago
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    i know a little.

  5. bahrom7893
    • 5 years ago
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    C' = 2 * 0.45x - 110 = 0

  6. heisenberg
    • 5 years ago
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    by taking the first derivative and setting it equal to 0 you can get the critical points

  7. bahrom7893
    • 5 years ago
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    0.9x - 110 = 0 .9x = 110

  8. heisenberg
    • 5 years ago
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    these are points where the function is at a max or minimum. since it is a parabola there should be only one, and hopefully it is a min (a second derivative test could confirm this).

  9. bahrom7893
    • 5 years ago
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    x = 122.22

  10. bahrom7893
    • 5 years ago
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    And basically whatever heisenberg said. Please fan us both if we helped. Thanks! =)

  11. heisenberg
    • 5 years ago
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    bahrom is exactly right, but what you get from this is the x coordinate of the critical point (min or max value). since the question wants the actual value of the function, plug this x value in the original equation: C(122.22)

  12. bahrom7893
    • 5 years ago
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    yup, good lol I missed the ending.. i just got back from college.. brains not functionin properly lol

  13. anonymous
    • 5 years ago
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    what happened to 10000?? you just kick that off??

  14. heisenberg
    • 5 years ago
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    when you take the derivative of a constant it is 0.

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