A catenoid is obtained by rotating the graph of the function $f(x)=a \cosh (x/a)$ around the $x$-axis.
Consider catenoids that satisfy the boundary condition $f(c)=f(-c)=r>0$. We have $a\cosh(c/a)=r$, and hence $$ \cosh (c/a)=(c/a)(r/c), $$ so there exists a number $R_0>0$, such that when $r/c>R_0$, there are exactly two catenoids spanned by the two coaxial circles with the same radius $r$.
My question is: which one of the two catenoids has larger surface area? This seems to be much more difficult than it appears to be. Indeed, I can compute the area explicitly: $$\pi a (a\sinh (2c/a)+2c) $$ but the two values of $a$ are very hard to characterise, so I find it very difficult to compare the areas.
See this link for a diagram of the two catenoids.