# Global extrema without numbers

1. Nov 27, 2009

### Razael

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

When you cough, your windpipe contracts. The speed, v, with which air comes out depends on the radius, r, of your windpipe. If R is the normal (rest) radius of your windpipe, then for r <_ R, the speed if given by v = a(R-r)r2 where a is a positive constant. What value of r maximizes v? What is the maximum speed? Show all work. Use the procedure for finding global extrema to justify that you have found the global maximum velocity.

2. Relevant equations

v = a(R-r)r2

3. The attempt at a solution

Derivative is 2ar(R-r) - ar2 (R is considered a constant, right?)

= 2arR - 2ar2 -ar2

= 2arR - 3ar2 = 0

2arR = 3ar2

2aR = 3ar

r = (2/3)R

Not too confident in that answer. For maximum speed I just plugged it into the original equation:

v = a((2/3)R)2((1/3)R)

Any help is appreciated, whether it be by pointing out a mistake or helping with the last part (justify answer).

Edit The interval would be 0 < r <_ R, right?

Last edited: Nov 27, 2009
2. Nov 27, 2009

### Hurkyl

Staff Emeritus
(You've assumed that r is nonzero in your calculuation. Do you know where?)

Not sure if this is right? Well, let's look at what you've proven:

If v = a(R-r)r², and a and R are constants w.r.t. r, then dv/dr = 0 if and only if r = (2/3) R. (Assuming r is nonzero)

Well, that wasn't what you were trying to show now, is it? You were trying to say something about extrema, not zeros of derivatives! You need to add something to your argument. It might be a minor thing, but it needs to be added.

P.S. what is "the procedure for finding global extrema"?

3. Nov 29, 2009

### Razael

Here?

2arR = 3ar2

2aR = 3ar

Now that I think about it, r could be 0 (I assumed breathing was occurring )

It said extrema so I assumed I'd be finding critical points. Am I going about that wrong?

Finding function values at all critical points and endpoints.

4. Nov 29, 2009

### Razael

Just to clarify, r <_ R means r is less or equal to R.

5. Nov 29, 2009

### Hurkyl

Staff Emeritus
Yep.

Mainly, I wanted to make sure you didn't mentally replace "find the extrema" with "find where the derivative is zero".

The procedure you cite here is a reasonable one for finding extrema. And you are executing the "find the critical points" bit correctly. But from your original post, I couldn't tell if you knew why you were finding the critical points, or if you were going to remember to check the endpoints.

6. Nov 29, 2009

### Razael

I assumed r wasn't 0 because doing so would make v 0 (which wouldn't be a maximum). So I ended up with two local maxes; (2/3)R and R (the right end of the interval).

So now I put these values into the original function and find which results in a higher value, correct?

Edit: Using interval-end R as a value of r results in a function value of 0. (2/3)R is all I'm left with.

7. Nov 29, 2009

### Hurkyl

Staff Emeritus
If that's what the theorem about global maxima that you're using says, then yes, that's what you do!

A minor digression: while R is a local extremum, you haven't shown it to be a maximum. In fact, it can't be a local maximum: between any two local maxima, there must be a local minimum!

8. Nov 29, 2009

### Razael

Good point. Both r=0 and r=R result in the function equaling 0, so I guess they would both be minimums while r=(2/3)R would be the maximum (so it'd look like a concave down parabola).

Thanks for the help