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 $0<\dfrac{r_{0}}{2}\leq r\leq r_{0}.$

Solution

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 $r_{0}.$

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 $20\%.$