# Thread: optimization and coughing - max-min word-problem

1. ## optimization and coughing - max-min word-problem

Got this problem on a test wrong, I misread it quite badly.

According to one model of coughing, the flow F (volume per unit time) of air through the windpipe during a cough is a function of the radius r of the windpipe, given by:

for , where k is a constant and r0 is the normal (non-coughing) radius.

a) Find the value of r that maximizes the flow F.
b) According to the same model, the velocity v of air through the windpipe during a cough is given by:

for . Find the value of r that maximizes the velocity v.

c) During a cough, the windpipe is constricted. According to parts (a) & (b), is that likely to assist or hinder the cough?

Lengthy problem. On my test I differentiated incorrectly, and that messed up my whole problem. How do I do this right?

2. ## Re: optimization and coughing - max-min word-problem

$F = k(r_0 \cdot r^4 - r^5)$

$\frac{dF}{dr} = k(4r_0 \cdot r^3 - 5r^4)$

$\frac{d^2F}{dr^2} = k(12r_0 \cdot r^2 - 20r^3) = 4kr^2(3r_0 - 5r)$

setting $\frac{dF}{dr} = 0$ ...

$4r_0 \cdot r^3 - 5r^4 = 0$

$r^3(4r_0 - 5r) = 0$

$r = \frac{4r_0}{5}$ ... maximum $F$ since $\frac{d^2F}{dr^2} < 0$ for this value of $r$

$v = \frac{k}{\pi}(r_0 \cdot r^2 - r^3)$

$\frac{dv}{dr} = \frac{k}{\pi}(2r_0 \cdot r - 3r^2)$

$\frac{d^2v}{dr^2} = \frac{k}{\pi}(2r_0 - 6r)$

setting $\frac{dv}{dr} = 0$ ...

$2r_0 \cdot r - 3r^2 = 0$

$r(2r_0 - 3r) = 0$

$r = \frac{2r_0}{3}$ ... maximum for $v$ since $\frac{d^2v}{dr^2} < 0$ for this value of $r$

3. ## Re: optimization and coughing - max-min word-problem

So, I'd say the that if the windpipe is constricted, the cough would most likely be hindered. But how does that relate to the problem (how do the answers influence the last part)?

4. ## Re: optimization and coughing - max-min word-problem

Since flow and velocity are a maximum when the windpipe is restricted, I would say it helps, rather than hinders the cough.

5. ## Re: optimization and coughing - max-min word-problem

now I see... have to watch my time from now on.