In open-channel hydraulics, a “most economical” rectangular channel section minimizes wetted perimeter for a given area (or maximizes discharge for a given perimeter). For such a rectangular channel, what proportional relationship between depth and width must be maintained to achieve hydraulic efficiency? State the condition clearly and choose the correct statement.

Introduction / Context:
In civil engineering water resources and irrigation design, the “most economical” (or hydraulically efficient) channel shape minimizes the wetted perimeter for a given flow area, reducing friction losses. For a rectangular channel, there is a well-known proportionality between depth and width that achieves this condition.

Given Data / Assumptions:

• Prismatic open channel with rectangular cross-section.
• Objective: minimize wetted perimeter P for a given area A.
• Steady, uniform flow; roughness constant; gravity drainage.

Concept / Approach:
The hydraulic radius R is A / P. For fixed A, maximizing R (equivalently minimizing P) yields the most economical section. For a rectangle of width b and depth y: A = b * y, P = b + 2y. Using calculus, differentiate P with respect to b (or y) under a constant A constraint to find the optimum proportion.

Step-by-Step Solution:
Let A = b * y = constant.Express b = A / y.Wetted perimeter P = b + 2y = (A / y) + 2y.Minimize P: dP/dy = d/dy[(A / y) + 2y] = (-A / y^2) + 2.Set derivative to zero: (-A / y^2) + 2 = 0.Solve for y: A / y^2 = 2 → A = 2y^2 → by = 2y^2 → b = 2y.Therefore, depth y = b / 2.

Verification / Alternative Check:
Compute hydraulic radius at optimum: R = A / P = (by) / (b + 2y) = (2y*y) / (2y + 2y) = (2y^2) / (4y) = y / 2, confirming standard textbook result for rectangular channels at economical section.

Why Other Options Are Wrong:
One-fourth of the width: y = b/4 gives larger P for same A, not minimum.Three times hydraulic radius: y = 3R is not the defining condition here; for the economical rectangle R = y/2.Hydraulic mean depth: this is another name for hydraulic radius in some contexts, not a direct proportion b–y.None of these: incorrect because a correct option exists.

Common Pitfalls:

• Confusing triangular or trapezoidal optimal proportions with the rectangular case.
• Maximizing area instead of minimizing wetted perimeter for a fixed area.
• Mixing up hydraulic radius R with depth y; for the economical rectangle R = y/2.

Final Answer:
depth equals half the width