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? 
 A: Use the equation $S = 6l^2$ and the fact that the surface area is $216~\text{cm}^2$ to solve for $l$, then use the equation 
$$24 = 3l^2~\frac{dl}{dt}$$
to solve for $dl/dt$.  Substitute for $l$ and $dl/dt$ to solve for $dS/dt$.
A: The quantity $\frac{V^2}{S^3}$ is the same for all spatial figures of the same shape. For instance, for  spheres it is $\frac{(4\pi R^3/3)^2}{(4 \pi R^2)^3}= \frac{1}{36 \pi}$, while for cubes it is $\frac{1}{6^3}$. Therefore
$$d\ \log  \frac{V^2}{S^3} = 0$$ or 
$$2 \frac{dV}{V} - 3 \frac{dS}{S}=0$$
From here we get 
$$d S = \frac{2}{3}\frac{S}{V} d V$$
Since it's about cubes, we have $V= (\frac{S}{6})^{3/2}$. Substitute and  get the result. 
A: 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:
$$V=S\cdot \frac{x}{6} \Rightarrow \frac{dV}{dt}=\frac{dS}{dt}\cdot \frac{x}{6}+S\cdot \frac{1}{6}\cdot \frac{dx}{dt}=24 \Rightarrow \\
\frac{dS}{dt}\cdot \frac{6}{6}+216\cdot \frac{1}{6}\cdot \frac{2}{9}=24 \Rightarrow \\
\frac{dS}{dt}=16. $$
A: Direct substitution is also possible:


*

*$V = \left( \frac{S}{6}\right)^{\frac{3}{2}}$ or  $S = 6V^{\frac{2}{3}} \Rightarrow$


$$\frac{dS}{dt}=6\frac{2}{3}V^{-\frac{1}{3}}\frac{dV}{dt} =4V^{-\frac{1}{3}}\frac{dV}{dt}$$


*

*$V^{-\frac{1}{3}} = \left( \frac{S}{6}\right)^{-\frac{1}{2}} = \frac{\sqrt{6}}{\sqrt{S}} \Rightarrow$


$$\frac{dS}{dt}=4\sqrt{6}\frac{1}{\sqrt{S}}\frac{dV}{dt}$$
