10 - Largest conical tent of given slant height

$V = \frac{1}{3} \pi r^2 h$
 

Where:
$h = s \sin \theta$

$r = s \cos \theta$
 

$V = \frac{1}{3} \pi s^3 \cos^2 \theta \sin \theta$

$V = \frac{1}{3} \pi s^3 (1 - \sin^2 \theta) \sin \theta$

$V = \frac{1}{3} \pi s^3 (\sin \theta - \sin^3 \theta)$

$\dfrac{dV}{d\theta} = \frac{1}{3}\pi s^3 (\cos \theta - 3\sin^2 \theta \cos \theta) = 0$

$\cos \theta - 3\sin^2 \theta \cos \theta = 0$

$1 - 3\sin^2 \theta = 0$

$\sin^2 \theta = 1/3$

$\sin \theta = \sqrt{1/3} = 1 / \sqrt{3}$
 

$h = s \sin \theta = \frac{1}{\sqrt{3}}\,s$

$r = s \cos \theta = \sqrt{\frac{2}{3}}\,s$           answer