Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 12

a.) Recall that the volume of the cube is $\text{Vol } = x^3$ where $x$ is the sides of the cube.

$
\begin{equation}
\begin{aligned}
\frac{dV}{dx} &= \frac{d}{dx} (x^3)\\
\\
\frac{dV}{dx} &= 3x^2

\end{aligned}
\end{equation}
$


when $x = 3$mm


$
\begin{equation}
\begin{aligned}
\frac{dV}{dx} &= 3(3)^2\\
\\
\frac{dV}{dx} &= 27 \frac{mm^3}{mm}

\end{aligned}
\end{equation}
$

$\displaystyle \frac{dV}{dx} = 27$ means that the volume is increasing at a rate of $\displaystyle 27 \frac{mm^3}{mm}$ when the length of the cube is 3$mm$.

b.) Recall that the surface area of the cube is equal to $A = 6s^2$ and the rate of change of the volume of the cube is...

$
\begin{equation}
\begin{aligned}
V'(s) &= 3s^2 && \text{;where } s^2 = \frac{A}{6}\\
\\
V'(s) &= 3 \left( \frac{A}{6}\right)\\
\\
V'(s) &= \frac{A}{2}
\end{aligned}
\end{equation}
$