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Find the absolute maximum for f(t) = -t^3 + 3t^2 + 400t + 5000

  1. Mar 5, 2012 #1

    939

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    1. The problem statement, all variables and given/known data

    Find the absolute maximum for f(t) = -t^3 + 3t^2 + 400t + 5000, t is between 6 and 20.


    2. Relevant equations

    f'(t) = -3t^2 + 6t + 400

    3. The attempt at a solution

    I know how to find this - when the function can be perfectly factored. Yet some problems occur here.

    f'(t) = -3t^2 + 6t + 400

    I normally factor to find minimum, but 400 divided by -3 = -133.33333~, meaning the answer will be approximate (technically) if I do that... Now the second derivative will come to something really easy, 6t+6... Can absolute maximums be found with the second derivative? If not, how can it be found here?
     
  2. jcsd
  3. Mar 5, 2012 #2

    Mark44

    Staff: Mentor

    First, find any local maxima by setting f'(t) = 0. For your problem, use the Quadratic Formula to find zeroes. The second derivative can be used to determine whether critical points are maxima, minima, or inflection points.

    A global max/min can occur and any of three places:
    a) local critical points (at which f'(t) = 0)
    b) points in the domain of f at which f' is undefined - not relevant for your problem
    c) endpoints of the domain - very relevant for your problem
     
  4. Mar 5, 2012 #3
    You're on the right track, but you need to set the derivative equal to zero to find a maximum. Even though it can't be easily factored, you can still use the quadratic formula to find your values of t. You'll use the second derivative to see if it's a maximum, minimum, or inflection point.

    Remember that on a bounded interval you need to check the bounds to see if those are maxima too.
     
  5. Mar 5, 2012 #4

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    Thanks to both of you. Yes, the way to solve it was with the quadratic formula, 100%. My mistake :).
     
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