Maximum area with Minkowski measures of perimeter

Abstract

The oldest competition for an optimal shape (area-maximizing) was won by the circle. But if the fixed perimeter is measured by the line integral of |dx| + |dy|, a square would win. Or if the boundary integral of max(|dx|, |dy|) is given, a diamond has maximum area. For any norm in R 2 , we show that when the integral of (dx, dy) around the boundary is prescribed, the area inside is maximized by a ball in the dual norm (rotated by π/2).This "isoperimetrix" was found by Busemann. For polyhedra it was described by Wulff in the theory of crystals. In our approach, the EulerLagrange equation for the support function of S has a particularly nice form. This has application to computing minimum cuts and maximum flows in a plane domain.

Maximum flows and minimum cuts in the plane

“…In [32] we prove that the unit ball in the dual norm (rotated by π/2) is isoperimetrically optimal. Here that ball is a circle or a diamond or a square.…”

“…We compute the minimum cuts in the three versions of the challenge problem on the unit square. We also mention an isoperimetric problem (with a different definition of perimeter) to which we return in a later paper [32].…”

A Generalization of the Convex Kakeya Problem

“…The length λ of the perimeter of K is equal to the integral over the support function [23] λ = 2π 0 h(u θ )dθ.…”

Restricted Cuts for Bisections in Solid Grids: A Proof via Polygons

“…Due to the well-known isoperimetrical problem using the L 1 -norm (see e.g. [10]) it follows that λ is a rectangular line if the boundary points of λ do not coincide with two of the opposing corners of R. If λ's boundary points coincide with two opposing corners of R it is easy to see that λ can be exchanged with a straight, corner, or staircase line since these lines have minimum length between the boundary points using the L 1 -norm.…”