Most economical trapezoid of area 40 m²

Reading of the side slope: "1:2 (H:V)" is taken as 1 horizontal to 2 vertical, so $z=0.5$.

Given data

$A=40$ m², $z=0.5$, $S_0=1/1500$, $C=50$.

Most economical trapezoidal section

Conditions: $R=\dfrac y2$ and $b=2y(\sqrt{1+z^2}-z)$.

$$A=(b+zy)y=y^2\left(2\sqrt{1+z^2}-z\right)=y^2\left(2\sqrt{1.25}-0.5\right)=1.7361\,y^2$$ $$y=\sqrt{\frac{40}{1.7361}}=4.800\ \text{m}$$ $$b=2y(\sqrt{1.25}-0.5)=2\times4.800\times0.6180=5.933\ \text{m}$$

Check: $A=(5.933+0.5\times4.800)\times4.800=40.0$ m².

Wetted perimeter $P=b+2y\sqrt{1+z^2}=16.666$ m; $R=A/P=2.400$ m $=y/2$ (✓).

Discharge

$$V=C\sqrt{RS_0}=50\sqrt{\frac{2.400}{1500}}=2.000\ \text{m/s}$$ $$Q=AV=40\times2.000=80.00\ \text{m}^3/\text{s}$$

Answer: depth $y=4.80$ m, bed width $b=5.93$ m, discharge $Q=80.0$ m³/s.