# Thread: Optimization problem

1. ## Optimization problem

Hi,

I hope someone can provide some clarification on the following "algorithm":

I'm wondering whether this algorithm would be more correct had it included the use of the second derivative test in order to confirm whether the critical values are at a max or min? I see a lot of examples of these optimization problems, and the majority of them are solved with the use of the second derivative test. I'm curious whether the second derivative test is actually necessary or has it just been used to further confirm whatever finding's one made about the max or min? Based on this algorithm, it appears that the second derivative test isn't necessary. I hope someone can explain why.

- Olivia

2. ## Re: Optimization problem

The "second derivative test" isn't used because that would require that the function have a second derivative, making the algorithm less general. In this case, the first derivative test is used to determine all points at which there might be a maximum or minimum, and there can only be a finite number of them, then the actual values are compared to see which are maximum or minimum.

3. ## Re: Optimization problem

but the algorithm isn't even suggesting to use the first derivative test, all it is saying is to plug the critical points into the functions and to then compare them. Is that really enough to confirm which point is a max or min within an interval?

4. ## Re: Optimization problem

Existence Theorem: A continuous function on a closed interval [a,b] has an absolute maximum and an absolute minimum on that interval

The algorithm mentioned above is used to determine these absolute min and max points

The first and second derivative tests are used to determine local maximum and local minimum points

5. ## Re: Optimization problem

'Critical points' are where the first derivative is 0 or does not exist. That is the first derivative test.

6. ## Re: Optimization problem

okay gotcha, thanks

7. ## Re: Optimization problem

Originally Posted by HallsofIvy
'Critical points' are where the first derivative is 0 or does not exist. That is the first derivative test.
no, that is not the first derivative test.

The first derivative test includes the following steps

1) Find the critical points by solving $f '(x)=0$
and then
2) test each such critical point to see whether it is a max,min or neither

In the above algorithm we only need to use 1) to find the critical points. We do not need the second step.
There is no need to determine whether those critical points are max,min or neither.
The algorithm will work and will produce the absolute max and min of f in [a,b]

8. ## Re: Optimization problem

The process of finding absolute extrema on a continuous, closed interval is sometimes called the "candidate" test.

The possible "candidates" for extrema on a closed, continuous interval $[a,b]$ are the endpoints, $f(a)$ and $f(b)$, and any values of $x \in (a,b)$ where $f'(x)=0$ or is undefined.

Evaluating each and comparing will yield the absolute extrema on the closed interval.

Local extrema w/in the closed interval that are not absolute will require using the first and/or second derivative tests for classification as a max or min.

9. ## Re: Optimization problem

Yeah, thanks for clarifying. It's referred to in my textbook as the "extreme value theorem". I figured out though

10. ## Re: Optimization problem

Originally Posted by Idea
no, that is not the first derivative test.

The first derivative test includes the following steps

1) Find the critical points by solving $f '(x)=0$
No, determine where f'(x)= 0 or f'(x) does not exist as I said before.

and then
2) test each such critical point to see whether it is a max,min or neither
also, if this is close, bounded interval, test the end points.

In the above algorithm we only need to use 1) to find the critical points. We do not need the second step.
There is no need to determine whether those critical points are max,min or neither.
The algorithm will work and will produce the absolute max and min of f in [a,b]