30. For what dimensions does an open bathtub, having the shape of a half-cylinder with surface area $S$, have the greatest capacity
30. Denoting the length of the bathtub by $l$, and the radius of the base by $r$, we can write

$$
S=\pi r^{2}+\pi r l, v=\frac{1}{2} \pi r^{2} l
$$

or $4 \pi v^{2}=\pi r^{2}\left(S-\pi r^{2}\right)^{2}$. Since the sum $\pi r^{2}+\left(S-\pi r^{2}\right)=S$ is a constant, $4 \pi v^{2}$, and therefore $v$, attains its maximum value when the condition $\frac{\pi r^{2}}{1}=\frac{S-\pi r^{2}}{2}$ is satisfied. From this, we get $r=\sqrt{\frac{S}{2 \pi}} ; l=2 r$.