Answer
a) $\frac{3}{2}s^{2}\csc\theta(\csc\theta-{\sqrt 3}\cot\theta)$
b) $55°$
c) $6s\left(h+\frac{s}{2\sqrt 2}\right)$
Work Step by Step
$S$ = $6sh-\frac{3}{2}s^{2}\cot\theta+3s^{2}\left(\frac{\sqrt 3}{2}\right)\csc\theta$
a)
$\frac{ds}{d\theta}$ = $\frac{3}{2}s^{2}\csc^{2}\theta-3s^{2}\left(\frac{\sqrt 3}{2}\right)\csc\theta\cot\theta$ = $\frac{3}{2}s^{2}\csc\theta(\csc\theta-{\sqrt 3}\cot\theta)$
b)
$\frac{ds}{d\theta}$ = $0$
$\csc\theta-{\sqrt 3}\cot\theta$ = $0$
$\cos\theta$ = $\frac{1}{\sqrt 3}$
The First Derivative Test shows that the minimum surface area occurs when $\theta$ = $\cos^{-1}\frac{1}{\sqrt 3}$ $\approx$ $55°$
c)
$\cos\theta$ = $\frac{1}{\sqrt 3}$
$\cot\theta$ = $\frac{1}{\sqrt 2}$
$\csc\theta$ = $\frac{\sqrt 3}{\sqrt 2}$
so the surface area is
$S$ = $6sh-\frac{3}{2}s^{2}\left(\frac{1}{\sqrt 2}\right)+3s^{2}\left(\frac{\sqrt 3}{2}\right)\left(\frac{\sqrt 3}{\sqrt 2}\right)$ = $6s\left(h+\frac{s}{2\sqrt 2}\right)$
