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, 6 months ago

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

Samuraiwarm Tsunayoshi - 6 years, 6 months ago

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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, 6 months ago

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