Most economical trapezoidal channel of area 50 m²

For the most economical trapezoidal section: half the top width equals one sloping side, and $R = y/2$.

Data

Side slope 1 : 2 (H : V), so horizontal per unit vertical $n = 0.5$; $S = 1/1500$; $A = 50$ m²; $C = 50$.

Condition and dimensions

$$\begin{aligned} \frac{b + 2ny}{2} &= y\sqrt{1+n^2} \\ b &= 2y\sqrt{1+0.25} - 2(0.5)y = (2.2361 - 1)y = 1.2361y \\ A &= (b + ny)y = (1.2361 + 0.5)y^2 = 1.7361y^2 \\ 50 &= 1.7361y^2 \;\Rightarrow\; y = 5.367\ \text{m} \\ b &= 1.2361 \times 5.367 = 6.634\ \text{m} \end{aligned}$$

Check: $A = (6.634 + 0.5 \times 5.367) \times 5.367 = 9.317 \times 5.367 = 50.0$ m². Wetted perimeter $P = 6.634 + 2 \times 5.367 \times 1.118 = 18.63$ m, $R = 50/18.63 = 2.683$ m $= y/2$.

Discharge (Chezy)

$$\begin{aligned} Q &= AC\sqrt{RS} = 50 \times 50 \times \sqrt{2.683 \times \frac{1}{1500}} \\ &= 2500 \times 0.04229 = 105.7\ \text{m}^3/\text{s} \end{aligned}$$

  |<------ T = 12.00 m ------->|
  \                            /
   \     y = 5.37 m     1H:2V /
    \________________________/
          b = 6.63 m

Answer: Bottom width ≈ 6.63 m, depth ≈ 5.37 m, discharge ≈ 105.7 m³/s (velocity ≈ 2.11 m/s).