En guete Rutsch! (A Swiss New Year's greeting)

Happy New Year, Otto! Thanks for a wonderful problem to start off $\textbf{2019}$ !

The volume of a cylinder $C$ in $\mathbb{R}^n$ has the form $V_C \propto L\cdot B^{\,n-1}$ , where $L$ is the "height"' of the cylinder, and $B^{\,n-1}$ is the volume of the $(n-1)$ -sphere. For this problem, $n = 2019$ , $L \leq b$ , and $a = r$ is the radius of the $(n-1)$ -sphere. Now, because $B^{\,n-1} \propto r^{\,n-1} = a^{\,n-1}$ , we get $V_C \propto \vert b \vert\times a^{n-1} = \vert b \vert \times a^{2018}$ . But since the cylinder has maximum volume and is inscribed in the unit $2019$ -sphere, i.e., $\sum_{i=1}^{2019} x_i^2 \leq 1$ , we also require that $a^2+b^2 = 1$ .

The problem boils down to finding $\vert a \vert$ and $\vert b \vert$ with $a^2 + b^2 = 1$ such that $\vert b \vert \times a^{2018}$ is maximized, to which we simply proceed with calculus. Without loss of generality, let us assume $a,b > 0$ :

$\frac{d}{db}V_C \propto \frac{d}{db} ba^{2018} = \frac{d}{db}b\left(\sqrt{1-b^2}\right)^{2018} = 0$ if $\vert b\vert = \frac{1}{\sqrt{1+2018}} = \frac{1}{\sqrt{2019}}$ . This gives $\vert a\vert = \frac{\sqrt{2018}}{\sqrt{2019}}$ . So the answer we seek is $\frac{2018}{a^2} = \frac{2018\times2019}{2018} = \boxed{2019}$ .