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

[939]

Homework Statement

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

Homework Equations

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

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?

 

[Mark44]

Homework Statement

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

Homework Equations

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

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?

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

 

[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.

 

[939]

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