anonymous
  • anonymous
how to find the minimum value of this quadratic equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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heisenberg
  • heisenberg
which quadratic function would that be?
anonymous
  • anonymous
C(x)=.045x^2-110x+10000
heisenberg
  • heisenberg
this seems like you'll want to use calculus. finding the max/min values using the first derivative test. do you know derivatives/calculus?

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anonymous
  • anonymous
i know a little.
bahrom7893
  • bahrom7893
C' = 2 * 0.45x - 110 = 0
heisenberg
  • heisenberg
by taking the first derivative and setting it equal to 0 you can get the critical points
bahrom7893
  • bahrom7893
0.9x - 110 = 0 .9x = 110
heisenberg
  • heisenberg
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).
bahrom7893
  • bahrom7893
x = 122.22
bahrom7893
  • bahrom7893
And basically whatever heisenberg said. Please fan us both if we helped. Thanks! =)
heisenberg
  • heisenberg
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)
bahrom7893
  • bahrom7893
yup, good lol I missed the ending.. i just got back from college.. brains not functionin properly lol
anonymous
  • anonymous
what happened to 10000?? you just kick that off??
heisenberg
  • heisenberg
when you take the derivative of a constant it is 0.

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