Economical triangular section hydraulic radius

Most economical triangular section

Let side slope be $z$ (H:V), depth $y$:

$$A=zy^2,\qquad P=2y\sqrt{1+z^2}$$

For a given area, $y^2=A/z$, so

$$P^2=4y^2(1+z^2)=\frac{4A(1+z^2)}{z}$$

Minimise $P$ with respect to $z$:

$$\frac{d}{dz}\left(\frac{1+z^2}{z}\right)=\frac{2z\cdot z-(1+z^2)}{z^2}=0\;\Rightarrow\;z^2=1\;\Rightarrow\;z=1$$

So the best triangular section has side slopes $45^\circ$ (right-angled vertex at the bottom):

$$A=y^2,\qquad P=2\sqrt2\,y,\qquad R_t=\frac{y^2}{2\sqrt2\,y}=\frac{y}{2\sqrt2}$$

Most economical rectangular section

$b=2y$: $A=2y^2$, $P=4y$, so

$$R_r=\frac{2y^2}{4y}=\frac y2$$

Comparison (at the same depth $y$)

$$\frac{R_t}{R_r}=\frac{y/(2\sqrt2)}{y/2}=\frac{1}{\sqrt2}$$ $$\boxed{R_t=\frac{1}{\sqrt2}R_r}$$

Hence the hydraulic radius of the most economical triangular section is $1/\sqrt2$ times that of the most economical rectangular section of the same depth.