The Figure of Maximum Area and Given Perimeter

Date: 06/02/98 at 16:42:11
From: Brad Morris
Subject: Calculus/geometry  max/min problem

Hi Math Doctor!

I am a math teacher at Friends Central School in Wynnewood PA.

In our Geometry book (USCMP) it states that the geometric figure of 
maximum area and given perimeter is a circle. In the margin it states 
that Calculus is required to prove it. I'm the Calculus teacher and 
our Geometry specialist asked me about it and I really couldn't come 
up with a proof. I need help!

Thanks.
Brad Morris

Date: 06/02/98 at 18:54:43
From: Doctor Anthony
Subject: Re: Calculus/geometry  max/min problem

We can start by first showing that for, say, a rectangular shape of 
given perimeter, the square gives the greatest area.

If the perimeter is P, let length be x, then width is (1/2)(P-2x) and 
area is:

   A = (1/2)x(P-2x)

     = Px/2 - x^2

   dA/dx = P/2 - 2x  = 0  for max or min   

So 2x = P/2  and  x = P/4.

Thus the square is the optimum shape for a 4-sided figure.

This can be extended to figures with more sides, and it is always true 
that a symmetrical shape will enclose the greatest area for a given 
perimeter.

If we have a circle with radius r the perimeter is 2*pi*r.

If this is also the perimeter of a square, then each side of the 
square has length 2*pi*r/4 = pi*r/2 and area of square is:

   (pi*r/2)(pi*r/2) = (pi/4)pi*r^2

                    = pi/4 * area of circle

and since pi/4 < 1 the area of the square is less than the area of 
the circle.

A similar exercise with a 6-sided figure gives:

   area of hexagon = sqrt(3)pi/6 * area of circle

                   =  0.9068 * area of circle

If we then go to an n-sided figure with perimeter P, the area of one 
of the basic triangles formed by joining the midpoint of the figure 
to the vertices is:

   (1/2)r * cos(pi/n) * P/n  

(here r = distance from centre point to a vertex), and with n such 
triangles the area is  (1/2)r * P * cos(pi/n).

This will tend to its greatest value as cos(pi/n) -> 1 which requires 
that n -> infinity. So we end up with a polygon of an infinite number 
of sides, which is a circle.
         
-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   

Date: 06/02/98 at 22:43:33
From: Doctor Pat
Subject: Re: Calculus/geometry  max/min problem

Brad,
  
This is not a calculus answer, but I think if I had to explain to a 
typical geometry student why the theorem was true, this would be 
my approach.

STEP 1: Given any convex polygon with a given perimeter P, we can 
increase the area if any two adjacent sides are not the same.  

To illustrate this we draw a random n-gon with not all sides 
congruent. If two adjacent sides, call them r and s, are not 
congruent, draw the chord joining their two non-adjacent vertices, 
call it b, and form a triangle. The rest of the n-gon's area is not 
affected by changing the two sides r and s. We can improve the area if 
we maximize the triangle with a base equal to the chord b, and two 
sides that have a sum of r+s. I think most Geometry classes have 
proven that this triangle is maximized when the triangle is isosceles.  
Repeating this around the sides, eventually they must all end up the 
same. What we have demonstrated is that of all polygons with N sides, 
the regular polygon has the highest ratio for area/perimeter.  

STEP 2: Given any regular polygon with perimeter P and N sides, A 
regular Polygon with N+1 sides and perimeter P has a larger area. 

To illustrate this we begin with a regular n-gon of any number of 
sides. We know that any such n-gon is circumscribable so we draw the 
circle around it. Remember that the area of the n-gon is given by 
A = .5*Perimeter*apothem, which can be rewritten as A/P = apothem/2. 
That is, for any given perimeter, the area/perimeter ratio is directly 
proportional to the size of the apothem. We now show any one of the 
triangles formed by two radii of the circumscribing circle and one 
side of the n-gon. The perpendicular bisector of this side of the 
n-gon forms the altitude of the triangle, and the apothem of the 
n-gon. As we reduce the size of the side form P/n to P/(n+1) the angle 
gets a little smaller, and the ratio of the apothem to the radius of 
the circumscribing circle (which is a little smaller also for the n+1 
gon) gets a little larger. We need to show that the apothem is 
actually larger than in the previous case by using one of the trig 
functions, probably the cosine. Thus we show that the perimeter has 
stayed constant, but the apothem has gotten bigger, indicating  that 
the area is larger.  

I would expect this to be received much better if we first took say an 
octagon or decagon and found the area and perimeter and walked through 
the verification of each step. If your kids are like mine, they will 
nod 100 times when I show a proof they don't quite understand, but if 
I confirm it with numbers, they not only know it's true, they actually 
believe it's true, which is more important for me.  

Good luck.

-Doctor Pat,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   

Date: 11/07/2000 at 09:54:53
From: George
Subject: Finding the largest area using the same perimeter

How do I prove that the shape with the largest area and with a 
perimeter (or circumference) of 1000 m is a circle? Also, where do I 
go once I have done this? Is there an extension?

Date: 11/07/2000 at 15:11:00
From: Doctor Rob
Subject: Re: Finding the largest area using the same perimeter

Thanks for writing to Ask Dr. Math, George.

Here is one way to proceed:

Step 1: Show that a polygon with a fixed perimeter and largest area 
must be convex. Do this by considering any vertex with an interior 
angle of more than 180 degrees, and constructing the diagonal 
connecting the two vertices adjacent. The diagonal and the two edges 
will form a triangle. Reflect that triangle in the line that is the 
diagonal extended, and you have a polygon with a larger area and the 
same perimeter.

Step 2: Show that a convex polygon with a fixed perimeter and largest 
area must have all sides of the same length. Do this by looking at two 
unequal adjacent sides, and the triangle formed by them and the 
diagonal connecting their two outside ends. Show that the isosceles 
triangle with the same base and perimeter has a larger area, so 
replacing the constructed triangle with the isosceles one will give a 
polygon with the same perimeter and a larger area.

Step 3: Show that a convex polygon with equal sides, fixed perimeter, 
and largest area, must have all its interior angles of the same 
measure. Do this by looking at two adjacent angles of unequal measure, 
and the three sides forming them. Connect a diagonal from the outside 
ends of the sides to form a quadrilateral. Show that the isosceles 
trapezoid formed by the diagonal and three sides of the same length as 
the sides of the polygon has the same perimeter and larger area than 
the quadrilateral constructed, so replacing the three given sides and 
two included angles with three sides of the same lengths and two equal 
included angles forms a polygon with the same perimeter and larger 
area. (The fact that, given the lengths of the sides, the 
quadrilateral with the largest area is cyclic, that is, all its 
vertices lie on a single circle, should help here.)

These three steps show that among all polygons with n sides and a 
fixed perimeter, the regular n-gon has the largest area.

Step 4. Show that the area K of a regular n-gon with fixed perimeter P 
increases as n increases. Each side has length P/n, so

   K = n*(P/n)^2*cot(Pi/n)/4
   K = (P^2/4*Pi)*(Pi/n)/tan(Pi/n)

so you are reduced to showing that x/tan(x) increases as x approaches 
0 from above (in fact, its limit is 1.)

The limiting case of a regular n-gon as n increases without bound is a 
circle, which thus is the solution to the problem.

There is another solution using the calculus of variations, which I 
won't elaborate on here.

Can you generalize to polyhedra and spheres?

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   

Date: 04/08/2003 at 18:06:16
From: Mark
Subject: Area of a triangle

What are the dimensions of a triangle with perimeter p that encloses 
the maximum area?

Date: 04/08/2003 at 21:31:06
From: Doctor Paul
Subject: Re: Area of a triangle

The maximum area occurs when the triangle is equilateral with each 
side having length p/3.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/