Let the width of a river be $a$ and the width of a canal be $b$. Canal is perpendicular to the river. What is the maximum length of a log that can be send from river to canal?
The result given in a book is $(a^{2/3} + b^{2/3})^{3/2}$.
I have found this exercise in the book of exercises by Boris Demidovič. It is known in eastern Europe to be the one of the best collections of problems.
edit: changed word to perpendicular, thanks Henry Lee.
What you have is the river of width $a$ (shown in red) met by a canal of width $b$ (shown in blue). For the log to be able to flow from blue to red it must be able to pivot freely about the meeting point (joint of red and blue lines). Imagine that the log starts off with an angle $\theta=0$ where it is fully in the blue region, and it is safely on the river when said angle reaches $\theta=\frac\pi2$. Come up with an expression that gives the maximum length of the log in terms of $a,b,\theta$ and the minimum value of this will be the maximum length of the log.
Hope this helps :)