Isosceles Triangle Maximizes Area?

Date: 09/11/2003 at 16:31:37
From: LaShawn
Subject: isosceles triangle

How can you show that among all triangles having a specified base 
and a specified perimeter, the isosceles triangle on that base has 
the largest area?

I know that 

  b + c =  p - a 

where p is the perimeter.  I also know that 

  A^2 = s(s-a)(s-b)(s-c) 

and that the area is a maximum when a = b = c.  But I don't know where
to go from there. 

Date: 09/12/2003 at 03:58:56
From: Doctor Floor
Subject: Re: isosceles triangle

Hi, LaShawn.

Thanks for your question.

Indeed we may use Heron's formula,

  A^2 = s(s-a)(s-b)(s-c)

here.  Let a be the base, which is fixed, and let 

  2t = b + c 

We may introduce the only variable x by b=t-x and c=t+x.

Substitution into Heron's formula gives:

  A^2 = s(s-a)(s-t-x)(s-t+x) 

      = s(s-a)((s-t)^2 - x^2)

Since x is the only variable, clearly

   s(s-a)((s-t)^2 - x^2) 

reaches a maximum when x=0. And x=0 gives b=c=t, which makes the
triangle isosceles. 

Here is a second, more geometric, way to see why the triangle will be
isosceles.

Let's fix base AB of a triangle. Then AC+BC is fixed, hence C lies on 
an ellipse with foci AB. Line AB is the major axis. The points on the 
ellipse at maximal distance of the major axis are on the minor axis. 
These are the points such that AC=BC. By A = 1/2 bh this shows that 
these points also give maximal area of ABC.

If you have more questions, just write back.

Best regards,

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