 QUESTION

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

$\Rightarrow$ Volume of a cube $= {x^3}$
$\Rightarrow$Surface area $= 6{x^2}$
$\Rightarrow $\dfrac{{dv}}{{dt}} = 8c{m^3} per second. Then by using the chain rule, We get \Rightarrow 8 = \dfrac{{dv}}{{dt}} = \dfrac{{d{x^3}}}{{dt}} \times \dfrac{{dx}}{{dx}} Or 8 = 3{x^2} \times \dfrac{{dx}}{{dt}} \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{8}{{3{x^2}}} ……………………………………………(1) Now, we will solve \dfrac{{ds}}{{dt}}, \Rightarrow \dfrac{{d\left( {6{x^2}} \right)}}{{dt}} = \dfrac{{d\left( {6{x^2}} \right)}}{{dt}} \times \dfrac{{dx}}{{dx}} \\ = \dfrac{{d\left( {6{x^2}} \right)}}{{dx}} \times \dfrac{{dx}}{{dt}} \\ = 12x \times \dfrac{{dx}}{{dt}} \\ We used chain rule to solve the above mentioned equation. \Rightarrow 12x \times \dfrac{{dx}}{{dt}} \Rightarrow 12x \times \dfrac{8}{{3{x^2}}} \Rightarrow \dfrac{{32}}{x} Thus using the given condition, When x = 12 cm \Rightarrow$\dfrac{{ds}}{{dt}}$ = $\dfrac{{32}}{{12}}c{m^2}$per second
$\Rightarrow$$\dfrac{8}{3}$ $c{m^2}$ per second is the right answer.
$\therefore$ Hence if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of $\dfrac{8}{3}$ $c{m^2}$per second.