My Math Forum Proving The Maximum Area of a Shape Regardless of How Many Sides It Has

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 December 30th, 2017, 04:09 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 660 Thanks: 87 Proving The Maximum Area of a Shape Regardless of How Many Sides It Has The greatest area of a quadrilateral given a fixed perimeter is a square. This seems to be true for an equilateral triangle. Is there a proof that given a fixed perimeter and fixed amount of sides, the area will be maximized when all the sides are equal regardless of how many sides there are?
 December 30th, 2017, 05:00 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics I'm a bit short on time atm but here is the idea. Suppose a $n$-polygon is specified as a list of $n+1$ vertices, $\{(x_0,y_0),\dotsc,(x_n,y_n)\}$ and let $\gamma_k$ denote the line segment between the $(k-1)^{\rm st}$ and $k^{\rm th}$ vertex. Then the area is given explicitly by the line integral $\sum_{k=1}^n \int_{\gamma_k} x \ dy$ Now, regard this as a function of $2(n+1)$-many variables and maximize it using standard multi-variable optimization techniques. Thanks from Maschke
 December 31st, 2017, 12:46 PM #3 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 For any figure more than 3 sides, it is important to include a requirement that all angles be equal. Example for 4 sides: a rhombus has all sides equal, but it can be squeezed to an area as close to 0 as one wants. Thanks from Country Boy
January 4th, 2018, 04:19 PM   #4
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 Originally Posted by Knowledgesearcher I got a lot of answers there.
Oh, I'm sure you did..

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