Question

The volume of a cube is increasing at the rate of $8 cm^3 /s$. How fast is the surface area increasing when the length of an edge is $12 cm$?

Answer 1

The correct answer is $\frac{8}{3} cm^2 /s.$
Let $x$ be the length of a side, $V$ be the volume, and $s$ be the surface area of the cube. Then, $V = x^3$ and $S = 6x^2$ where $x$ is a function of time $t$.
It is given that $\frac{dv}{dt}=8cm^3/s$
Then, by using the chain rule, we have:
$∴ 8=\frac{dv}{dt}=\frac{d}{dt}(x^3)=\frac{dx}{dt}(x^3).\frac{dx}{dt}=3x^2.\frac{dx}{dt}$
$\frac{dx}{dt}=\frac{8}{3x^2}....(1)$
Now, $\frac{ds}{dt}=\frac{d}{dt}(6x^2)=\frac{d}{dt}(6x^2).\frac{dx}{dt}$ ...[By chain rule]
$=12x.\frac{dx}{dt}=12x.\frac{dx}{dt}=12x.(\frac{8}{3x^2})=\frac{32}{x}$
$\frac{ds}{dt}=\frac{32}{12}cm^2/s=\frac{8}{3}cm^2/s.$
Thus, when $x = 12 cm,$ Hence, if the length of the edge of the cube is $12 cm$, then the surface area is increasing at the rate of $\frac{8}{3} cm^2 /s.$