Isoperimetric Problem

Prove that (in Euclidean two-space) given a finite perimeter, the closed curve enclosing maximum area is a circle.


Before reading the rest of the solution, first read about Green's Theorem and area.

We begin with the area of a closed curve derived from Green's theorem and the formula for arclength: A=12xdyydx=12xyydxA = \frac{1}{2} \oint{xdy-ydx} = \frac{1}{2} \oint{xy'-y} dx

S=1+y2dx.S = \int \sqrt{1+{y'}^{2}} dx.

Using Lagrange multipliers, we consider the expression A=[12(xyy)+λ1+y2]dxA = \oint \left[\frac{1}{2} (xy'-y) + \lambda \sqrt{1+{y'}^{2}} \right] dx which is like the action functional in classical mechanics.

If you want to learn about how Lagrange developed the multiplier method, click here.

Treating [12(xyy)+λ1+y2]dx\left[\frac{1}{2} (xy'-y) + \lambda \sqrt{1+{y'}^{2}} \right] dx as the "Lagrangian", the Euler-Lagrange equation is ddx(12x+λy1+y2)+12=0\frac{d}{dx} \left(\frac{1}{2} x + \lambda \frac{y'}{\sqrt{1+{y'}^{2}}}\right) + \frac{1}{2}=0 or ddx(λy1+y2)=1.\frac{d}{dx} \left(\frac{\lambda y'}{\sqrt{1+{y'}^{2}}} \right) = -1.

Since λ\lambda is a scalar, we note that the above expression can be written y(1+y2)3/2=1λ\frac{y''}{{(1+{y'}^{2})}^{3/2}} = \frac{-1}{\lambda} which states that the curvature of this maximizing curve is constant. This would either be a circle or a straight line; however, straight lines are not closed and can possess infinite arclength. So given a finite perimeter, the circle encloses the maximum area.

Check out my set Classic Demonstrations.

Note by Steven Zheng
6 years, 2 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

Is it about isoperimetric inequality? I couldn't prove the theorem though.

Samuraiwarm Tsunayoshi - 6 years, 2 months ago

Log in to reply

You can solve this using integration and the vector calculus theorem A=12xdyydx.A = \frac { 1 }{ 2 } \oint { xdy-ydx } . However there is a purely geometric proof if you change the question to polygons. The calculus of variation method is true for general closed curves in R2{\mathbb{R} }^{2}.

Steven Zheng - 6 years, 2 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...