04-05 Stiffness and strength of timber beam

$k = bd^3$

Where:
$b = D \cos \theta$

$d = D \sin \theta$
 

$k = D^4 \cos \theta \, \sin^3 \theta$

$\dfrac{dk}{d\theta} = D^4 (3\cos^2 \theta \, \sin^2 \theta - \sin^4 \theta) = 0$

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

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

$\tan^2 \theta = 3$

$\tan \theta = \sqrt{3}$

$\theta = 60^\circ$
 

$b = D \cos 60^\circ = \frac{1}{2} D$

$d = D \sin 60^\circ = \frac{1}{2} \sqrt{3} D$
 

$\text{depth} = \sqrt{3} \times \text{breadth}$           answer

$S = bd^2$

Where:
$b = D \cos \theta$

$d = D \sin \theta$
 

$S = D^3 \cos \theta \sin^2 \theta$

$S = D^3 \cos \theta (1 - \cos^2 \theta)$

$S = D^3 (\cos \theta - \cos^3 \theta)$

$\dfrac{dS}{d\theta} = D^3 (-\sin \theta + 3\cos^2 \theta \sin \theta) = 0$

$-1 + 3\cos^2 \theta = 0$

$\cos^2 \theta = \frac{1}{3}$

$\cos \theta = \frac{1}{\sqrt{3}}$
 

$b = D \cos \theta = \frac{1}{\sqrt{3}}D$

$d = D \sin \theta = \dfrac{1}{\sqrt{3}} \sqrt{2} \, D$
 

$\text{depth } = \sqrt{2} \times \text{ breadth}$           answer