Question

The volume of a cube is increasing at the rate of $8c{m}^3/s$. If the length of an edge is $12cm$, then the surface area is increasing at the rate of

• A. $\frac{4}{3}c{m}^2/s$
• B. $\frac{8}{3}c{m}^2/s$
• C. $\frac{2}{3}c{m}^2/s$
• D. None of these

Answer 1

Answer: (B)

Let $x$ be the length of a side (edge), $V$ be the volume and $S$ be the surface area of the cube.
$V=x^3$ and $S=6x^2$
( $\because$ cube has six square faces, each of side $x$ )
where, $x$ is a function of timet. It is given that $\frac{dV}{dt}=8c{m}^3/s$.
Then, by using the chain rule, we have
$8=\frac{dV}{dt}=\frac{d}{dt}\left(x^3\right)=3x^2\cdot \frac{dx}{dt}\Rightarrow \frac{dx}{dt}=\frac{8}{3x^2}.....$(i)
Now, $\frac{dS}{dt}=\frac{d}{dt}\left(6x^2\right)=12x\frac{dx}{dt}=12x\left(\frac{8}{3x^2}\right)=\frac{32}{x}$
[from Eq. (i)]
Thus, when $x=12cm,\frac{dS}{dt}=\frac{32}{12}c{m}^2/s=\frac{8}{3}c{m}^2/s$.
Hence, if the length of the edge of a cube is $12cm$, then the surface area is increasing at the rate of $\frac{8}{3}c{m}^2/s$.