Derive most economical rectangular section

A most economical (best hydraulic) section carries the maximum discharge for a given area, slope and roughness. Since $Q=\dfrac{1}{n}AR^{2/3}S^{1/2}$, for a given $A$ the discharge is maximum when the hydraulic radius $R=A/P$ is maximum, i.e. when the wetted perimeter $P$ is minimum.

Derivation

Rectangular channel of bed width $b$ and depth $y$:

$$A=by\;\Rightarrow\;b=\frac Ay,\qquad P=b+2y=\frac Ay+2y$$

For minimum $P$ with $A$ constant:

$$\frac{dP}{dy}=-\frac{A}{y^2}+2=0\;\Rightarrow\;A=2y^2$$ $$\frac{d^2P}{dy^2}=\frac{2A}{y^3}>0\quad\text{(minimum)}$$

Hence $by=2y^2$:

$$\boxed{b=2y}$$

Hydraulic radius:

$$R=\frac AP=\frac{2y^2}{2y+2y}=\frac y2$$ $$\boxed{y=\frac b2,\qquad R=\frac y2}$$

Conclusion: the most economical rectangular section has bed width equal to twice the depth (depth = half the width) and hydraulic radius equal to half the depth. Equivalently, it is half of a square of side $b$ cut across the middle.

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 |   y = b/2      |  y
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        b = 2y