Perimeter of 1000m

Date: 07/13/99 at 14:59:31
From: Louise Macdonald
Subject: A shape with the maximum area and a perimeter of 1000m

We have to find a shape with a perimeter of 1000m and the maximum 
area. I have tried to work this out but need to know if a circle 
would have the biggest area. I have already looked at rectangles, 
triangles, and polygons and have noticed that the equilateral shape 
had the maximum area of each group of shapes.

Thank you.

Date: 07/13/99 at 15:09:48
From: Doctor Anthony
Subject: Re: A shape with the maximum area and a perimeter of 1000m

Consider various regular polygons with n = number of sides. You will 
have a constant perimeter (= 1000m) and you want an expression for the 
area in terms of n. When n = 3 you have an equilateral triangle with 
each side 333.33 metres.  

   area = (1/2)333.33^2 sin(60) = 48112.52 m^2

For a square of side 250 m 

   area = 62500 m^2

From here you require a general formula. If a polygon has n sides, the
angle subtended by any side at the centre is 360/n. Suppose the 
polygon is inscribed in a circle of radius r; then the area of any 
triangle formed by radii from the centre to the edges of a side of 
the polygon is

  (1/2)r^2.sin(360/n)

and the area of the whole polygon is (n/2)r^2.sin(360/n)

where the perimeter is given by 2n.r.sin(180/n) = 1000

                                 n.r.sin(180/n) = 500

So you have two formulae to work with:

   r = 500/[n.sin(180/n)] .....(1)

and area   

   A = (n/2)r^2.sin(360/n)  .......(2)

Check with n = 3  r = 500/[3.sin(60)] = 192.4500897
                 
                  A = (3/2)(192.4500897)^2 sin(120) = 48112.52 

and this agrees with the value we found earlier.

Combining the two formulae we get the area of a regular polygon of n 
sides having a perimeter of 1000m:

            (500)^2
   A =  -----------------
          n.tan(pi/n)

which can be changed to    (500^2/n).cot(pi/n)

                         = (500^2/n).tan(pi/2 - pi/n)

or if you want to express it in degrees

           A = (500^2/n).tan(90 - 180/n)

Plot a graph of A against n and show that A increases with an increase 
in n, but at a rate that decreases as n increases. The obvious limit 
is when you have n -> infinity, in which case you have a circle. Find 
the area of the circle. This will be the limit toward which A is 
heading as n is increased.

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