Depth for maximum discharge, triangular channel

Reading of the figure: the section is a closed equilateral triangle conduit (both sides at $60^\circ$ to the horizontal) with the base horizontal at the bottom and apex at the top, partly filled to depth $h$. (For an open V-shaped section the discharge would increase continuously with depth, so a maximum exists only for the closed apex-up section.) Let the side of the triangle be $a$ and its height $H=\dfrac{\sqrt3}{2}a$.

        /\
       /  \         apex up
      / ~~ \  h     water of depth h
     /______\
        a

Geometry for depth $h$

Width of water surface at height $h$: $T=a\left(1-\dfrac hH\right)$.

$$A=ah-\frac{a}{2H}h^2=a\left(h-\frac{h^2}{2H}\right)$$

Wetted perimeter: base + the two sloping sides below the surface ($h/\sin60^\circ=2h/\sqrt3$ each):

$$P=a+\frac{4h}{\sqrt3}$$

Let $u=h/H$. Then $\dfrac{4h}{\sqrt3}=\dfrac{4Hu}{\sqrt3}=2au$:

$$A=aH\left(u-\frac{u^2}{2}\right),\qquad P=a(1+2u)$$

Maximum discharge

$Q=KAR^{2/3}=KA^{5/3}P^{-2/3}$. For maximum $Q$: $\dfrac{dQ}{dh}=0$:

$$5P\frac{dA}{dh}=2A\frac{dP}{dh}$$ $$\frac{dA}{dh}=a(1-u),\qquad \frac{dP}{dh}=\frac{2a}{H}$$ $$5\,a(1+2u)\,a(1-u)=2\,aH\left(u-\frac{u^2}{2}\right)\frac{2a}{H}$$ $$5(1+u-2u^2)=4u-2u^2\;\Rightarrow\;8u^2-u-5=0$$ $$u=\frac{1+\sqrt{1+160}}{16}=\frac{1+12.689}{16}=0.8555$$ $$h=uH=0.8555\,H=0.7409\,a$$

Answer: the discharge is maximum when the depth is about 0.856 of the height of the triangle ($h\approx0.856H\approx0.741a$, where $a$ is the side).