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

**Solution**

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: \[A = \frac{1}{2} \oint{xdy-ydx} = \frac{1}{2} \oint{xy'-y} dx\]

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

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

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

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

Since \(\lambda\) is a scalar, we note that the above expression can be written \[\frac{y''}{{(1+{y'}^{2})}^{3/2}} = \frac{-1}{\lambda}\] 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.

## Comments

Sort by:

TopNewestIs it about isoperimetric inequality? I couldn't prove the theorem though. – Samuraiwarm Tsunayoshi · 2 years, 5 months ago

Log in to reply

– Steven Zheng · 2 years, 5 months ago

You can solve this using integration and the vector calculus theorem \[A = \frac { 1 }{ 2 } \oint { xdy-ydx } .\] 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 \({\mathbb{R} }^{2}\).Log in to reply