Answer
See graph and explanations.
Work Step by Step
a. Given the velocity equation $v=c(r_0-r)r^2=c(r_0r^2-r^3), \frac{r_0}{2}\leq r\leq r_0$, let its first derivative be zero; we have $v'=c(2r_0r-3r^2)=0$, which gives $r=\frac{2}{3}r_0$. Check, at this point, $v''=c(2r_0-6r)=c(2r_0-6(\frac{2}{3}r_0))=-2cr_0\lt0$. Thus, we know the region is concave down with a maximum.
b. For $r_0=0.5, c=1$, the equation becomes $v=0.5r^2-r^3$. Graph the function as shown and we can find a maximum at $r=\frac{1}{3}=\frac{2}{3}(0.5)=\frac{2}{3}(r_0)$
