When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied by a contraction of the trachea, making a narrower channel for the expelled air to flow through. For a given amount of air to escape in a fixed time, it must move faster through the narrower channel than the wider one. The greater the velocity of the airstream, the greater the force on the foreign object. X rays show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough. According to a mathematical model of coughing, the velocity $v$ of the airstream is related to the radius r of the trachea by the equation
$$v(\mathrm{r})=\mathrm{k}\left(\mathrm{r}_{0}-\mathrm{r}\right) \mathrm{r}^{2} \quad \frac{1}{2} \mathrm{r}_{0} \leqslant \mathrm{r} \leqslant \mathrm{r}_{0}$$
where $\mathrm{k}$ is a constant and $\mathrm{r}_{0}$ is the normal radius of the trachea. The restriction on $\mathrm{r}$ is due to the fact that the tracheal wall stiff- ens under pressure and a contraction greater than $\frac{1}{2} \mathrm{r}_{0}$ is prevented (otherwise the person would suffocate).
(a) Determine the value of $r$ in the interval $\left[\frac{1}{2} r_{0}, r_{0}\right]$ at which $v$ has an absolute maximum. How does this compare with experimental evidence?
(b) What is the absolute maximum value of $v$ on the interval?
(c) Sketch the graph of $v$ on the interval $\left[0, r_{0}\right] .$