Question

The volume of cube is increasing at a constant rate. Prove that the increase in surface area varies inversely as the length of edge of the cube.

Answer 1

Let the side of cube be $x$ unit
$\therefore$ Volume of cube $\left(V\right)=x^3$
On differentiating both side w.r.t. $t$ we get
$\frac{dV}{dt}=3x^2\frac{dx}{dt}=k$ [constant ]
$\Rightarrow \frac{dx}{dt}=\frac{k}{3x^2}....\left(i\right)$
Also surface area of cube $S=6x^2$
On differentiating w.r.t.r, we get
$\frac{dS}{dt}=12x.\frac{dx}{dt}$
$\Rightarrow \frac{dS}{dt}=12x.\frac{k}{3x^2}$ [using eq (i) ]
$\Rightarrow \frac{dS}{st}=\frac{12k}{3x}=4\left(\frac{k}{x}\right)$
$\Rightarrow \frac{dS}{dt}\alpha \frac{1}{x}$
hence the surface area of the cube varies inversely as the length of the side.