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.

(N/A) Let the side of a cube be $x$ units.
Volume of cube $V = x^{3}$.
Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 3x^{2} \frac{dx}{dt} = k$ (where $k$ is a constant).
$\Rightarrow \frac{dx}{dt} = \frac{k}{3x^{2}} \dots (i)$.
Surface area of the cube $S = 6x^{2}$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = 12x \cdot \frac{dx}{dt}$.
Substituting the value of $\frac{dx}{dt}$ from equation $(i)$:
$\frac{dS}{dt} = 12x \cdot \left( \frac{k}{3x^{2}} \right) = \frac{4k}{x}$.
Since $4k$ is a constant,$\frac{dS}{dt} \propto \frac{1}{x}$.
Thus,the rate of increase of the surface area varies inversely as the length of the side $x$.