The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

Let the side of a cube be $x$ unit.

Therefore the volume of cube $(V) = {x^3}$

On differentiating both side w.r.t. $t$, we get
$\frac{{dV}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}} = k$ [constant ]
$\Rightarrow$ $\frac{{dx}}{{dt}} = \frac{k}{{3{x^2}}}$ …..(i)

Also, surface area of cube, $S = 6{x^2}$

On differentiating w.r.t. $t$, we get
$\frac{{dS}}{{dt}} = 12x \cdot \frac{{dx}}{{dt}}$

$\Rightarrow$ $\frac{{dS}}{{dt}} = 12x \cdot \frac{k}{{3{x^2}}}$ [using Eq. (i)]

$\Rightarrow$ $\frac{{dS}}{{dt}} = \frac{{12k}}{{3x}} = 4\left( {\frac{k}{x}} \right)$

$\Rightarrow$ $\frac{{dS}}{{dt}} \propto \frac{1}{x}$

Hence, the surface area of the cube varies inversely as the length of the side.