Most economical trapezoidal channel section (maximum discharge) For the most economical (maximum discharge) trapezoidal channel section, which of the following conditions hold true simultaneously?

Introduction / Context:
The “most economical” open-channel section maximizes discharge (or minimizes wetted perimeter) for a given flow area, reducing friction losses and construction cost. For trapezoidal channels (widely used in irrigation and drainage), elegant geometric conditions characterize this optimum.

Given Data / Assumptions:

• Prismatic trapezoidal channel: bed width b, side slope z (horizontal:vertical), depth y.
• Uniform, steady, subcritical flow; resistance described by Manning or Chezy.
• Objective: minimize wetted perimeter P for given area A to maximize A/P (hence hydraulic radius R).

Concept / Approach:

For a given A, discharge Q ∝ R^(2/3) A^(5/3) (Manning). Maximizing Q reduces to maximizing R = A/P at fixed A, i.e., minimizing P. Using calculus of variations or standard derivations yields relations among b, y, and z for the optimum trapezoid.

Step-by-Step Solution:

Verification / Alternative check:

For the most economical triangle (b = 0), the side slope satisfies z = 1/√2; for a nonzero b the above relationships collapse consistently and reproduce classic textbook results such as R = y/2.

Why Other Options Are Wrong:

Each of (a), (b), and (c) is individually correct, so any single choice alone is incomplete. The comprehensive answer is that all stated geometric conditions are satisfied at the optimum.

Common Pitfalls:

Confusing hydraulic depth (A/T) with hydraulic radius (A/P); forgetting that “most economical” refers to geometry, not necessarily cost-only factors like lining roughness or freeboard.

Final Answer:

all the above