A circle has the largest possible area for given perimeter. Can you formally prove it.?

Question

A circle has the largest possible area for given perimeter. Can you formally prove it.?

Answer 1

1. It's easy to show that the curve capturing the largest area must be convex (otherwise, we can decrease the perimeter and increase the area of the curve).



2. It's easy to show that no polygon could be such curve: for any polygon of degree N, we can construct a polygon of degree N+1 with same perimeter but greater area. The way to do it:



consider any vertice of the polygon A. consider two more vertices on A's left and on A's right, let's call them B and C.



so, we have ABC triangle, AB+AC are contributing towards the perimeter, and BC is the base



If we build a trapezoid BCDE with the same base BC, D and E two new vertices to replace A, and BD=DE=EC, it's easy to show such trapezoid will have larger area than ABC triangle.



then since for a polygon with N vertices, polygon with N+1 is better, we see that the best polygon has an infinite amount of vertices = circle.



3. However for closed curves that aren't polygons... such as ellipses, for example.... let me think on this one :)



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Here's what we could try.



Lemma 1. for a given perimeter and given N, a regular polygon with N vertices will have larger area than any irregular polygon with N vertices (not gonna prove, but wouldn't be hard, argument similar to #2 above, vertice-by-vertice).



Let's now consider a closed curve C, some polygon inscribed in it, and some polygon in which the curve is inscribed. So, polygon on inside, polygon on outside. let's call them I and O.



it's clear that I's area S(I) < S(C), and perimeter P(I)
similarly, S(O)>S(C), and P(O)>P(C)



Lemma 2. For any closed curve C and any number d, no matter how small, there exist I and O of degree > N such



S(O)-S(C)


granted, N may have to get quite large for small d's. There will however exist a calculable N(d) dependence.



again, im not gonna formally prove this lemma.





what i'm trying to do is sandwhich the area and perimeter of the curve between two polygons, with which it's easier to work.



so now suppose we choose some N=3 and draw the I and O (triangles). If I and O are regular polygons, choose N=4, then N=5 etc. If I and O stay regular, N will run to infinity, and this will mean C is indeed a circle.



If for some N I and O are not regular, make new polygons I' and O' of the same degree and perimeter that are regular. By Lemma 1,



S(I')>S(I), and S(O')>S(O)



if S(I') > S(O), then we can draw a C' that's sandwhiched between I' and O', with the same perimeter and C, whose area is larger.



If not, note that



S(O') = S(O) + x, for some x. By Lemma 2, we can choose d to be smaller than x for some large N', and let N' be a multiple of N.



So now we can construct polygons I'' and O'', where for sure



S(I'') > S(O)



and thus the area of C'' sandwhiched between I'' and O'' will be greater than area of C





i hope you can follow that rabble....

Answer 2

Think you have solved your own problem. It's obv not a polygon or any concave closed curve, so draw an axis of symnetry through it. Now ..and this the good bit.. make a triangle on one half using a point on the perimeter. If you max the area of this triangle it corresponds to the max area of the section. Area triangle= 1/2basexheightxsin(angle subtended at perimeter)..max value=when angle=pi/2.

Can you guess what shape has a closed curve which has a straight side and any enclosed angle of 90deg at its perimeter..and the argument is symnetrical for the other section.

Answer 3

Let a given closed figure F represent the figure with the maximum area for its area. Draw a line through it. Then flip over either side and rejoin. If it creates a non-convex figure, then by using tangent lines a convex figure can be drawn with a better area-to-perimeter ratio can be found. Hence the closed figure F must have the property that it remains convex regardless of this "slice, flip, and rejoin" operation. This means that the angles made with the figure F must be the same at both ends of the slice. Let figure F sit on the x-axis as a tangent line, the point of tangency designated as the origin (0,0), and we draw lines radiating from (0,0), intersecting figure F, beginning with angle = 90 relative to the x-axis, and sweeping down to angle = 0. We can define a vector field in this manner, so that we have a series of nested figures which tangents correspond to this vector field. Since we already know that nested circles meets this criteria (proof by means of right triangles with the diameter as the hypotenuse), it excludes the possibility of nested closed figures of other than circles.

Answer 4

Perimeter, P = 2pi(r)

r = P/(2pi)



Area, A = pi(r)^2

A = pi(P/(2pi))^2

A = piP^2/[4(pi)^2]

A = P^2/[4pi]



As the perimeter has a given value and 4 and pi are both constants too, hence A has a constant value too.



The area is directly proportional to the square to the perimeter.

Answer 5

We want to maximize the area

A = pi (r)^2 ...........................(1)

Let its perimeter = B

B = 2 pi r

r = B/(2 pi)

From (1)

A = pi (B^2/4 (pi)^2)

A = 1/4 [(B^2)/(pi)]= constant

Ther is a unique value of A for a given value of B

Its a maximum since its derivative = 0

Answer 6

Every circle with the same perimeter has the same area.