Maximum-area cylinder inscribed in a sphere ($h/r=\sqrt{5}-1,A_{\max }=\pi R^2(1+\sqrt{5})$)

For a closed right circular cylinder with radius $r>0$, height $h>0$, and total surface area $A=2\pi r(r+h)$, inscription in a sphere of radius $R$ gives $r^2+h^2/4=R^2$. The Lagrange multiplier equations imply $h^2+2rh-4r^2=0$, so the positive aspect ratio is $h/r=\sqrt{5}-1$. The dimensions are $r^2=R^2(5+\sqrt{5})/10$ and $h^2=2R^2(5-\sqrt{5})/5$.

There is also an eigenvalue proof of global maximality. Set $w=h/2$, so the constraint is $r^2+w^2=R^2$, and write $A/(2\pi )=r^2+2rw$. This quadratic form has symmetric matrix $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ with largest eigenvalue $(1+\sqrt{5})/2$, the golden ratio. Its maximizing eigenvector lies in the positive quadrant, yielding the displayed area and aspect ratio.