Answer
a. $h=\frac{5r}{2}$ or $r=\frac{2h}{5}$
b. $-\frac{125}{144\pi}\approx-0.276ft/min$
Work Step by Step
a. Using the figure provided by the Exercise, we have $\frac{h}{10}=\frac{r}{4}$ (using similar triangles); thus, $h=\frac{5r}{2}$ or $r=\frac{2h}{5}$
b. With $V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi (\frac{2h}{5})^2h=(\frac{4\pi}{75})h^3$, we have $\frac{dV}{dt}=(\frac{4\pi}{25})h^2\frac{dh}{dt}$
Given $\frac{dV}{dt}=-5ft^3/min$ and $h=6ft$, we have $-5=(\frac{4\pi}{25})6^2\frac{dh}{dt}$ which gives $\frac{dh}{dt}=-\frac{125}{144\pi}\approx-0.276ft/min$
