Coughing is caused by increased pressure in the lungs and is accompanied by a decrease in the diameter of the windpipe. See Figure 20. The radius \(r\) of the windpipe decreases with increased pressure \(p\) according to the formula \(r_{0}-r=cp,\) where \(r_{0}\) is the radius of the windpipe when there is no difference in pressure and \(c\) is a positive constant. The volume \(V\) of air flowing through the windpipe is \[ V=kpr^{4} \]
where \(k\) is a constant. Find the radius \(r\) that allows the most air to flow through the windpipe. Restrict \(r\) so that \(0<\dfrac{r_{0}}{2}\leq r\leq r_{0}.\)
Since \(p=\) \(\dfrac{r_{0}-r}{c},\) we can express \(V\) as a function of \(r\): \[ V=V(r) =k\!\left( \dfrac{r_{0}-r}{c}\right) r^{4}=\dfrac{k r_{0}}{c} r^{4}-\dfrac{k}{c} r^{5} \qquad \dfrac{r_{0}}{2}\leq r\leq r_{0} \]
Now we find the absolute maximum of \(V\) on the interval \(\left[\dfrac{r_{0}}{2},r_{0}\right].\) \[ V^\prime (r) =\dfrac{4k\,r_{0}}{c}\,r^{3}-\dfrac{5k}{c}\,r^{4}=\dfrac{k}{c}\,r^{3}(4r_{0}-5r) \]
The only critical number in the interval \(\left( \dfrac{r_{0}}{2}, r_{0}\right)\) is \(r=\dfrac{4r_{0}}{5}.\)
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We evaluate \(V\) at the critical number and at the endpoints, \(\dfrac{r_{0}}{2}\) and \(r_{0}.\)
| \({r}\) | \({V(r) =k\left( \dfrac{r_{0}-r}{c}\right) r^{4}}\) |
|---|---|
| \(\dfrac{r_{0}}{2}\) | \(k\left( \dfrac{r_{0}-\dfrac{r_{0}}{2}}{c}\right) \left( \dfrac{r_{0}}{2}\right) ^{\!\!4}=\dfrac{k\,r_{0}^{5}}{32c}\) |
| \(\dfrac{4r_{0}}{5}\) | \(k\left( \dfrac{r_{0}-\dfrac{4r_{0}}{5}}{c}\right) \left( \dfrac{4r_{0}}{5}\right) ^{\!\!4}=\dfrac{k}{c}\cdot \dfrac{ 4^{4}\,r_{0}^{5}}{5^{5}}=\dfrac{256\,k\,r_{0}^{5}}{3125c}\) |
| \(r_{0}\) | \(0\) |
The largest of these three values is \(\dfrac{256\,kr_{0}^{5}}{3125c}.\) So, the maximum air flow occurs when the radius of the windpipe is \(\dfrac{4r_{0}}{5},\) that is, when the windpipe contracts by \(20\%.\)