Related rate of a melting cube's side length

Start with the volume formula for a cube, $V = s^3$, where both $V$ and $s$ are functions of time $t$. Differentiate both sides with respect to $t$ using the chain rule: $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$. This equation relates the rate of change of volume to the rate of change of side length at any instant.

Substitute the known values $\frac{dV}{dt} = -2$ (volume decreasing) and $s = 4$ into the equation: $-2 = 3(4)^2 \frac{ds}{dt} = 48 \frac{ds}{dt}$. Solving for $\frac{ds}{dt}$ gives $\frac{ds}{dt} = -\frac{2}{48} = -\frac{1}{24}$ cm/min, meaning the side length is decreasing at a rate of $\frac{1}{24}$ cm per minute.