the volume of the expanding cube is increasing at the rate of $24 cm^3/min$, how fast is the surface area increasing when surface area is $216cm^2$? If the volume of the expanding cube is increasing at the rate of $24 cm^3/min$, how fast is the surface area increasing when surface area is $216cm^2$?
My Approach :
$V=l^3$
$$\dfrac {dV}{dt}=3l^2\dfrac {dl}{dt}$$
$$24=3l^2\dfrac {dl}{dt}$$
Also, 
$S=6l^2$
$$\dfrac {dS}{dt}=12l.\dfrac {dl}{dt}$$
How do I proceed? 

The volume: $V=x^3$. The surface: $S=6x^2$. Given the surface is $216$, we can find the side:
$$6x^2=216 \Rightarrow x=6.$$
Given the volume increases at the rate $24$, we can find at what rate the side is increases:
$$\frac{dV}{dt}=3x^2\cdot \frac{dx}{dt}=24 \Rightarrow \frac{dx}{dt}=\frac{24}{3\cdot 6^2}=\frac29.$$
Given the volume and surface equation, we can differentiate the volume:
$$V=S\cdot \frac{x}{6} \Rightarrow \frac{dV}{dt}=\frac{dS}{dt}\cdot \frac{x}{6}+S\cdot \frac{1}{6}\cdot \frac{dx}{dt}=24 \Rightarrow \\
\frac{dS}{dt}\cdot \frac{6}{6}+216\cdot \frac{1}{6}\cdot \frac{2}{9}=24 \Rightarrow \\
\frac{dS}{dt}=16. $$
