Question

The volume of a cube is increasing at the rate of $18{\text{cm}}^3$ per second. When the edge of the cube is $12cm,$ then the rate in $c{m}^2/s$ at which the surface area of the cube increases, is

Answer 1

Answer: 6

Let, $x$ be the length of an edge, $V$ be the volume and $S$ be the surface area of the cube.
$\frac{dV}{dt}=18,V=x^3\Rightarrow \frac{dV}{dt}=18=3x^2\frac{dx}{dt}\Rightarrow \frac{dx}{dt}=\frac{6}{x^2}\dots \left(i\right)$
$S=6x^2$
$\Rightarrow \frac{dS}{dt}=12x\frac{dx}{dt}\dots \left[\text{From}\left(i\right)\right]$
$\Rightarrow \frac{dS}{dt}=12x\frac{6}{x^2}$
$\Rightarrow \frac{dS}{dt}=12\times \frac{6}{x}=6\left[\because \text{Given }x=12\right]$
$\therefore \frac{dS}{dt}=6{\text{cm}}^{\text{2}}\text{/sec}$