Answer
$-\frac{5}{72\pi} in/min$ (decreasing)
$-\frac{10}{3} in^2/min$ (decreasing)
Work Step by Step
Step 1. We are given $2R_0=8, R_0=4in, \frac{dV}{dt}=-10in^3/min$
Step 2. The thickness $z$ of the ice gives a total radius of $r=4+z$
Step 4. Total volume $V=\frac{4}{3}\pi r^3$ and $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi r^2\frac{dz}{dt}$
Step 5. Plug-in the numbers; we have $-10=4\pi (4+2)^2\frac{dz}{dt}$, which gives $\frac{dz}{dt}=-\frac{5}{2\pi(36)}=-\frac{5}{72\pi} in/min$ (decreasing)
Step 6. Surface area $A=4\pi r^2$, $\frac{dA}{dt}=8\pi r\frac{dr}{dt}=8\pi (4+2)(-\frac{5}{72\pi} )=-\frac{10}{3} in^2/min$ (decreasing)
