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

1. Mar 5, 2012

939

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. Mar 5, 2012

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

3. Mar 5, 2012

louiswins

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.

4. Mar 5, 2012

939

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