Efficient rectangular section and triangular channel depth

Condition for the most efficient rectangular section

For a rectangular channel of width $b$ and depth $y$: $A = by$, $P = b + 2y = \dfrac{A}{y} + 2y$. For minimum $P$ at constant $A$:

$$\frac{dP}{dy} = -\frac{A}{y^2} + 2 = 0 \;\Rightarrow\; A = 2y^2 \;\Rightarrow\; b = 2y$$

So the width is twice the depth ($b = 2y$), and $R = \dfrac{2y^2}{4y} = \dfrac{y}{2}$.

Normal depth of triangular channel

Data: side slope $m = 2$ (2H : 1V), $S = 0.0005$, $Q = 0.5$ m³/s, $n = 0.04$.

$$\begin{aligned} A &= my^2 = 2y^2 \\ P &= 2y\sqrt{1+m^2} = 2y\sqrt{5} = 4.472y \\ R &= \frac{2y^2}{4.472y} = 0.4472y \end{aligned}$$

Manning's equation:

$$\begin{aligned} Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\ 0.5 &= \frac{1}{0.04}(2y^2)(0.4472y)^{2/3}(0.0005)^{1/2} \\ 0.5 &= 25 \times 2 \times 0.5848 \times 0.02236\, y^{8/3} = 0.6538\,y^{8/3} \\ y^{8/3} &= 0.7648 \;\Rightarrow\; y = 0.904\ \text{m} \end{aligned}$$

Check: $A = 2(0.904)^2 = 1.636$ m², $R = 0.404$ m, $Q = 25 \times 1.636 \times 0.404^{2/3} \times 0.02236 = 0.50$ m³/s.

Answer: Normal depth $y_n \approx 0.90$ m.