The first thing we are going to do is identify the volume and surface of the cube and their respective derivatives or rate of change
[tex]\begin{gathered} V\to\text{volume} \\ S\to\text{surface} \\ l=\text{side of a square} \end{gathered}[/tex][tex]\begin{gathered} V=l^3\to(1) \\ \frac{dV}{dt}=3l^2\frac{dl}{dt}\to(2) \end{gathered}[/tex][tex]\begin{gathered} S=6l^2\to(3) \\ \frac{dS}{dt}=12\cdot l\cdot\frac{dl}{dt}\to(4) \end{gathered}[/tex]From the exercise we know that:
[tex]\begin{gathered} \frac{dV}{dt}=300\frac{\operatorname{mm}^3}{s}\to(5) \\ 3l^2\frac{dl}{dt}=300\frac{\operatorname{mm}^3}{s}\to(2)=(5) \\ \frac{dl}{dt}=\frac{300}{3l^2}\frac{\operatorname{mm}^3}{s}\to(6) \end{gathered}[/tex]The exercise asks us to calculate the rate of change of the surface (4) so we substitute the differential of length (6) in (4)
[tex]\begin{gathered} \frac{dS}{dt}=12\cdot l\cdot(\frac{300}{3l^2}\frac{\operatorname{mm}^3}{s}) \\ \frac{dS}{dt}=\frac{1200}{l}\frac{\operatorname{mm}}{s} \end{gathered}[/tex]what is the rate of change of the cube’s surface area when its edges are 50 mm long?
[tex]\begin{gathered} l=50\operatorname{mm} \\ \frac{dS}{dt}=\frac{1200}{50\operatorname{mm}}\frac{\operatorname{mm}^3}{s} \\ \frac{dS}{dt}=24\frac{\operatorname{mm}^2}{s} \end{gathered}[/tex]