The strongest beam cut from a log

Strength. $h^2=d^2-w^2$, so $S(w)=w(d^2-w^2)=d^2w-w^3$ for $0<w<d$.

Maximum. $S'(w)=d^2-3w^2=0\iff w=\dfrac d{\sqrt3}$. Since $S'>0$ on $\left(0,\frac d{\sqrt3}\right)$ and $S'<0$ on $\left(\frac d{\sqrt3},d\right)$, it is the global maximum on $(0,d)$.

Dimensions.
$$w=\frac d{\sqrt3}\approx0.577\,d,\qquad h=\sqrt{d^2-\frac{d^2}3}=d\sqrt{\frac23}\approx0.816\,d.$$
Thus $\dfrac hw=\sqrt2$.

Remark. The strongest beam is not the one of largest area (that would be the square, $w=h$). Tall beams resist bending better, which is why floor joists are laid on edge.