The isoperimetric problem is the problem of finding the closed plane curve with a given perimeter that encloses the greatest area. Below, I will use the calculus of variations in a proof of the long-known solution, the circle.
We let our enclosed region be D, and the enclosing curve ∂D, with orientation on ∂D chosen so that D is to it's left (counterclockwise). We thus want to minimize subject to the constraint . Note that these are quite different kinds of integrals, and cannot be combined directly via the multiplier method of including constraints.
However, using Green's theorem, which says that over a plane region D with (counterclockwise oriented) boundary ∂D we have:
By choosing , and using the fact that the area of D is , we obtain
So now we can combine the minimized quantity and the constraint using a multiplier. As both are integral constraints, the multiplier is a constant λ, so parametrizing our curve ∂D as (x(t),y(t)), so
and we want to find the extrema of
The integrand is , so we have the Euler-Lagrange equation for x:
Now, , so , and
Similarly for y:
So we have and
Squaring and summing these, we obtain:
This is the equation of a circle with radius λ and and center (C2,C1). From our original constraint , we obtain λ=p/2π.