how to find the minimum value of this quadratic equation?

Question

heisenberg · Answer

which quadratic function would that be?

anonymous · Answer

C(x)=.045x^2-110x+10000

heisenberg · Answer

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

anonymous · Answer

i know a little.

bahrom7893 · Answer

C' = 2 * 0.45x - 110 = 0

heisenberg · Answer

by taking the first derivative and setting it equal to 0 you can get the critical points

bahrom7893 · Answer

0.9x - 110 = 0
.9x = 110

heisenberg · Answer

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 · Answer

x = 122.22

bahrom7893 · Answer

And basically whatever heisenberg said. Please fan us both if we helped. Thanks! =)

heisenberg · Answer

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 · Answer

yup, good lol I missed the ending.. i just got back from college.. brains not functionin properly lol

anonymous · Answer

what happened to 10000?? you just kick that off??

heisenberg · Answer

when you take the derivative of a constant it is 0.