---
title: "Derive most economical rectangular section"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:most-economical-rectangular-section-derive"
source: https://www.bctengineer.com/q/semester_4/hydraulics/most-economical-rectangular-section-derive
site: BCT Engineering
---

# Derive most economical rectangular section

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2075 Bhadra, 3 marks
- Video: [most economical rectangular channel section derivation](https://www.youtube.com/watch?v=yZlUdToqX2o) — e Tution
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/most-economical-rectangular-section-derive
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Derive the expression for most economical rectangular section.

## Short answer

For fixed area, minimizing wetted perimeter gives b = 2y, so depth is half the width. The hydraulic radius becomes R = y/2, representing a half-square section.

## Answer

A **most economical (best hydraulic) section** carries the maximum discharge for a given area, slope and roughness. Since $Q=\dfrac{1}{n}AR^{2/3}S^{1/2}$, for a given $A$ the discharge is maximum when the hydraulic radius $R=A/P$ is maximum, i.e. when the wetted perimeter $P$ is minimum.

#### Derivation
Rectangular channel of bed width $b$ and depth $y$:
$$
A=by\;\Rightarrow\;b=\frac Ay,\qquad P=b+2y=\frac Ay+2y
$$
For minimum $P$ with $A$ constant:
$$
\frac{dP}{dy}=-\frac{A}{y^2}+2=0\;\Rightarrow\;A=2y^2
$$
$$
\frac{d^2P}{dy^2}=\frac{2A}{y^3}>0\quad\text{(minimum)}
$$
Hence $by=2y^2$:
$$
\boxed{b=2y}
$$
Hydraulic radius:
$$
R=\frac AP=\frac{2y^2}{2y+2y}=\frac y2
$$
$$
\boxed{y=\frac b2,\qquad R=\frac y2}
$$
**Conclusion:** the most economical rectangular section has bed width equal to twice the depth (depth = half the width) and hydraulic radius equal to half the depth. Equivalently, it is half of a square of side $b$ cut across the middle.

```text
 ~~~~~~~~~~~~~~~~~~
 |                |
 |   y = b/2      |  y
 |________________|
        b = 2y
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a given area, slope and roughness, discharge is maximum when:
   A. the wetted perimeter is maximum
   B. the top width is minimum
   C. the wetted perimeter is minimum (hydraulic radius maximum)
   D. the depth is equal to the width
   Answer: C. Q = (1/n) A R^(2/3) S^(1/2) increases with R = A/P, so for fixed A it is largest when P is smallest. (Key point: Condition for best section)

2. With A = by constant, P = A/y + 2y. Setting dP/dy = 0 gives:
   A. A = y^2
   B. A = 2y^2, hence b = 2y
   C. A = 4y^2, hence b = 4y
   D. A = y^2/2, hence b = y/2
   Answer: B. dP/dy = -A/y^2 + 2 = 0 gives A = 2y^2, and since A = by, b = 2y. (Key point: Minimising P)

3. For the most economical rectangular section, the hydraulic radius is:
   A. R = y
   B. R = y/4
   C. R = 2y
   D. R = y/2
   Answer: D. R = A/P = 2y^2/(4y) = y/2. (Key point: Hydraulic radius of best rectangle)

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