# Isoperimetric Problem

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

Solution

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.

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.

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Note by Steven Zheng
3 years, 7 months ago

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

- 3 years, 7 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 $${\mathbb{R} }^{2}$$.

- 3 years, 7 months ago