The volume of a cube is increasing at the rate of $8c{m}^3/s$. How fast is the surface area increasing when the length of an edge is $12cm$?

Let $x$ be the length of a side, $V$ be the volume, and $S$ be the surface area of the cube.

Then, $V=x^3$ and $S=6x^2$

It is given that $\frac{dV}{dt}=8c{m}^3/s$.

Then, by using the chain rule, we have:

$\therefore 8=\frac{dV}{dt}=\frac{d}{dt}\left(x^3\right)\cdot \frac{d}{dx}=3x^2\cdot \frac{dx}{dt}$

$\Rightarrow \frac{dx}{dt}=\frac{8}{3x^2}.........\left(1\right)$

Now, $\frac{dS}{dt}=\frac{d}{dt}\left(6x^2\right)\cdot \frac{d}{dx}=\left(12x\right)\cdot \frac{dx}{dt}$ [By chain rule]

$=12x\cdot \frac{dx}{dt}=12x\cdot \left(\frac{8}{3x^2}\right)=\frac{32}{x}$

Thus, when $x=12$ cm, $\frac{dS}{dt}=\frac{32}{12}c{m}^2/s=\frac{8}{3}c{m}^2/s.$

Hence, if the length of the edge of the cube is $12$ cm, then the surface area is increasing at the rate of $\frac{8}{3}c{m}^2/s$.