Maximise a complete globe on a closed cylindrical base

The source surface-area model is

$\begin{aligned}&36.5\\ & =4\pi(5r/4)^2+2\pi r^2+2\pi rh.\end{aligned}$

Multiplying by four gives $146=33\pi r^2+8\pi rh$, hence $h=(146-33\pi r^2)/(8\pi r)$. A positive base requires $r>0$ and $r^2<146/(33\pi)$.

The complete volume is

$\begin{aligned}&V\\ & =\frac43\pi(5r/4)^3+\pi r^2h\\ & =\frac{125}{48}\pi r^3+\frac{73}{4}r-\frac{33}{8}\pi r^3\\ & =\frac{73}{4}r-\frac{73}{48}\pi r^3.\end{aligned}$

Set $V^{\prime}=73/4-73\pi r^2/16=0$ to obtain $r^2=4/\pi$. Reject the negative root, giving $r=2/\sqrt\pi$ m. It satisfies the positive-height constraint. Since $V^{\prime\prime}=-73\pi r/8<0$, this is the maximum.

$V_{\max}=(73/4)(2/\sqrt\pi)-(73\pi/48)(2/\sqrt\pi)^3=73/(2\sqrt\pi)-73/(6\sqrt\pi)=73/(3\sqrt\pi)\text{ m}^3$.