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