For basic solution, see
node 2360.
Solution by Chain Rule
$V = \frac{4}{3}\pi r^3$
$r = \left( \dfrac{3V}{4\pi} \right)^{1/3}$
when r = 12 ft
$V = \frac{4}{3}\pi (12^3)$
$V = 2304 \pi ~ \text{ft}^3$
$A = 4\pi r^2$
$A = 4\pi \left( \dfrac{3V}{4\pi} \right)^{2/3}$
$\dfrac{dA}{dt} = 4\pi \cdot \dfrac{d}{dt} \left[ \left( \dfrac{3V}{4\pi} \right)^{2/3} \right]_{V = 2304 \pi} \cdot \dfrac{dV}{dt}$
Using the derivative function of calculator
$\dfrac{dA}{dt} = 4\pi \cdot (0.0133) \cdot 2$
$\dfrac{dA}{dt} = 0.3333 ~ \text{ft}^2/\text{min}$ ← answer
Solution by Volume Flow Rate of Hydraulics
$V = \frac{4}{3}\pi r^3$
$V = \frac{1}{3}(4\pi r^2)r$
$V = \frac{1}{3}Ar$
$\dfrac{dV}{dt} = \dfrac{1}{3}\left( A \dfrac{dr}{dt} + r \dfrac{dA}{dt} \right)$
Note that dr/dt is a velocity, and from Fluid Mechanics, Av = Q :
$Q = \dfrac{1}{3}\left( Q + r \dfrac{dA}{dt} \right)$
When r = 12 ft
$2 = \dfrac{1}{3}\left( 2 + 12 \cdot \dfrac{dA}{dt} \right)$
$\dfrac{dA}{dt} = \dfrac{2(3) - 2}{12}$
$\dfrac{dA}{dt} = 0.3333 ~ \text{ft}^2\text{/min}$ ← answer
