Answer
A $1$-cm increase in $r$ leads to a greater increase in $V$ than a $1$-cm increase in $h$.
Work Step by Step
1. Estimate the increase of $V$ if there is a $1$-cm increase in $r$, that is $\Delta r = 1$.
$\Delta V \approx \frac{{\partial V}}{{\partial r}}\Delta r \approx \frac{{2\pi }}{3}rh\Delta r$
For $r = h = 12$, we get
$\Delta V \simeq \frac{{2\pi }}{3}\cdot12\cdot12\cdot1 = 96\pi $
2. Estimate the increase of $V$ if there is a $1$-cm increase in $h$, that is $\Delta h = 1$.
$\Delta V \approx \frac{{\partial V}}{{\partial h}}\Delta h \approx \frac{\pi }{3}{r^2}\Delta h$
For $r = h = 12$, we get
$\Delta V \simeq \frac{\pi }{3}\cdot{12^2}\cdot1 = 48\pi $
Comparing these two results we conclude that a $1$-cm increase in $r$ leads to a greater increase in $V$ than a $1$-cm increase in $h$.
