Most economical rectangular channel (minimum perimeter for given area) For a rectangular open channel section that is hydraulically most economical (minimum wetted perimeter for a given area), the width b must be:

Introduction / Context:
“Most economical” sections minimize the wetted perimeter for a given flow area, maximizing hydraulic radius and reducing friction losses. For rectangular channels, a simple geometric proportion yields the optimal shape for efficient conveyance.

Given Data / Assumptions:

• Rectangular cross-section of width b and depth y.
• Target: minimize wetted perimeter P for a given area A = b * y.
• Steady, uniform design context (Manning/Chezy).

Concept / Approach:
For a rectangle: P = b + 2y. With A fixed, use calculus (or Lagrange multipliers) to minimize P. The result is b = 2y, which maximizes hydraulic radius R = A / P and thus reduces head loss for given discharge.

Step-by-Step Solution:

Verification / Alternative check:
At b = 2y, hydraulic radius R = A / P simplifies to y/2, which is maximal for the rectangular family under fixed A.

Why Other Options Are Wrong:

• b = y or b = y/2 do not minimize P for fixed area; they lead to larger wetted perimeter and lower R.
• “None of these” is incorrect because a known optimum exists.

Common Pitfalls:
Confusing the result with the triangular or trapezoidal optima; forgetting that this condition pertains to rectangular sections only.

Final Answer:
twice the depth of flow