Answer
(a) $v$ has an absolute maximum when $r = \frac{2}{3}~r_0$
This value corresponds with the experimental evidence.
(b) $v(\frac{2}{3}r_0) = \frac{4~k~r_0^3}{27}$
(c) We can see a sketch of the graph below.
Work Step by Step
(a) $v(r) = k(r_0-r)~r^2$
We can find the absolute maximum of $v$ in the interval $[\frac{1}{2}r_0,r_0]$
We can find the values of $r$ where $v'(r) = 0$:
$v(r) = k(r_0-r)~r^2$
$v(r) = kr_0~r^2-kr^3$
$v'(r) = 2kr_0~r-3kr^2 = 0$
$2kr_0~r-3kr^2 = 0$
$kr~(2r_0-3r) = 0$
$r = 0$ or $r = \frac{2}{3}~r_0$
When $r = \frac{1}{2}r_0$:
$v(\frac{1}{2}r_0) = k(r_0-\frac{1}{2}r_0)~(\frac{1}{2}r_0)^2 = \frac{k~r_0^3}{8}$
When $r = \frac{2}{3}r_0$:
$v(\frac{2}{3}r_0) = k(r_0-\frac{2}{3}r_0)~(\frac{2}{3}r_0)^2 = \frac{4~k~r_0^3}{27}$
When $r = r_0$:
$v(r_0) = k(r_0-r_0)~(r_0)^2 = 0$
$v$ has an absolute maximum when $r = \frac{2}{3}~r_0$
This value corresponds with the experimental evidence: "X rays show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough."
(b) When $r = \frac{2}{3}r_0$:
$v(\frac{2}{3}r_0) = k(r_0-\frac{2}{3}r_0)~(\frac{2}{3}r_0)^2 = \frac{4~k~r_0^3}{27}$
The absolute maximum value of $v$ on the interval is $~~v(\frac{2}{3}r_0) = \frac{4~k~r_0^3}{27}$
(c) We can see a sketch of the graph below.
