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

Let us assume that at any instant of time $t$, the edge of the cube is $x$, surface area be $S$ and the volume be $y$ then
$V = {x^3}$ and $S = 6{x^2}$ …(i)

Differentiating (i) w.r.t. $t$, we get
$\Rightarrow \cfrac{{dV}}{{dt}} = 3{x^2}\cfrac{{dx}}{{dt}}$ …(ii)

and $\cfrac{{dS}}{{dt}} = 6\left( {2x} \right)\cfrac{{dx}}{{dt}}$ …(iii)

$\cfrac{{dV}}{{dt}} = 8{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}}$ (Given)$\Rightarrow 3{x^2}\cfrac{{dx}}{{dt}} = 8{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}}$ (using (ii))

$\Rightarrow 3{\left( {12{\rm{cm}}} \right)^2}\cfrac{{dx}}{{dt}} = 8{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}}$
$\Rightarrow \cfrac{{dx}}{{dt}} = \cfrac{8}{{432}}{\rm{cm/sec}} = \cfrac{1}{{54}}{\rm{cm/sec}}$

Substituting this value of $\cfrac{{dx}}{{dt}}$ in (iii), we get
$\cfrac{{dS}}{{dt}} = 12\left( {12{\rm{cm}}} \right)\left( {\cfrac{1}{{54}}{\rm{cm/sec}}} \right)$

$= \cfrac{8}{3}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/sec}}$
Therefore the rate of increase of surface area $= \cfrac{8}{3}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/sec}}$ .