= Solution
Maximize the total <surface area> $A=2\pi r(r+h)$ with the <constraint> $g=r^2+h^2/4=1$. The <Lagrange multiplier> equations are
$$
4\pi r+2\pi h=2\lambda r,\qquad2\pi r=\lambda h/2.
$$
From the second, $\lambda=4\pi r/h$. Substitute into the first and multiply by $h/(2\pi)$ to obtain $2rh+h^2=4r^2$. Thus the aspect ratio $q=h/r$ satisfies $q^2+2q-4=0$. Since $q>0$,
$$
\boxed{\frac hr=\sqrt5-1.}
$$
This is the <maximum-area cylinder inscribed in a sphere>. The dimensions and area, although not needed to specify the ratio, are
$$
r^2=\frac{5+\sqrt5}{10},\qquad h^2=\frac{2(5-\sqrt5)}5,\qquad A_{\max}=\pi(1+\sqrt5).
$$
There is only one positive stationary ratio. At the degenerate boundaries $r=0$ and $h=0$, the limiting areas are zero and $2\pi$, both below this value, confirming the maximum.
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