Question 1201003
<pre>

Tutor greenestamps above assumed that a rectangle of maximum area inscribed in
a circle is a square.  If you are not a calculus student, and/or your teacher
says that can be assumed, indeed his solution is acceptable.  Below I have
shown the solution in case that assumption is not permitted.

{{{drawing(400,400,-4,4,-4,4,graph(400,400,-4,4,-4,4), circle(0,0,3),
line(-2.5,sqrt(11)/2,2.5,sqrt(11)/2), 
line(-2.5,-sqrt(11)/2,2.5,-sqrt(11)/2),
line(-2.5,sqrt(11)/2,-2.5,-sqrt(11)/2),
line(2.5,sqrt(11)/2,2.5,-sqrt(11)/2),
locate(2.5+.1,sqrt(11)/2+.2,"(x,y)")

  )}}}

The area of the rectangle is given by A = (2x)(2y) = 4xy
We want to show that the rectangle is a square, that 
the length = 2x = the width 2y, or x = y and, as greenestamps 
essentially stated:

{{{x^2+y^2}}}{{{""=""}}}{{{9}}}
{{{x^2+x^2}}}{{{""=""}}}{{{9}}}
{{{2x*2}}}{{{""=""}}}{{{9}}}
{{{x^2}}}{{{""=""}}}{{{9/2}}}
{{{x}}}{{{""=""}}}{{{sqrt(9/2)}}}
{{{x}}}{{{""=""}}}{{{3/sqrt(2)}}}
Rationalizing the denominator
{{{x}}}{{{""=""}}}{{{expr(3/sqrt(2))expr(sqrt(2)/sqrt(2))}}}
{{{x}}}{{{""=""}}}{{{(3sqrt(2))/2}}}
So the length and with are both 2x = 2y = {{{3sqrt(2)}}}

----------------------------------------------------------------------

However in case you are required to prove that the rectangle is a square:
 
We will take the point (x,y) in QI so x and y are both positive.

{{{A}}}{{{""=""}}}{{{4xy}}}

We want to show that x = y so that the dimensions, 2x by 2y are the same.

Squaring both sides and differentiating explicitly avoids messy 
square roots and fraction exponents:

{{{A^2}}}{{{""=""}}}{{{16x^2y^2}}}
{{{2A*expr(dA/dx)}}}{{{""=""}}}{{{16(x^2*2y*expr(dy/dx)+y^2*2x)}}}
{{{2A*expr(dA/dx)}}}{{{""=""}}}{{{32x^2y*expr(dy/dx)+32xy^2)}}}
{{{A*expr(dA/dx)}}}{{{""=""}}}{{{16x^2y*expr(dy/dx)+16xy^2)}}}

Since A = 4xy, divide the left side by A and the right side by 4xy

{{{dA/dx}}}{{{""=""}}}{{{4x*expr(dy/dx)+4y)}}}

We need {{{dy/dx}}}

{{{x^2+y^2}}}{{{""=""}}}{{{9}}}
{{{2x+2y*expr(dy/dx)}}}{{{""=""}}}{{{0}}}
{{{2y*expr(dy/dx)}}}{{{""=""}}}{{{-2x}}}
{{{dy/dx}}}{{{""=""}}}{{{(-2x)/(2y)}}}
{{{dy/dx}}}{{{""=""}}}{{{-x/y}}}

Substituting,

{{{dA/dx}}}{{{""=""}}}{{{4x*expr(-x/y)+4y}}}

We set that equal to 0

{{{4x*expr(-x/y)+4y}}}{{{""=""}}}{{{0}}}
{{{-4x^2/y^""}}}{{{""=""}}}{{{-4y}}}
{{{-4x^2}}}{{{""=""}}}{{{-4y^2}}}
{{{x^2}}}{{{""=""}}}{{{y^2}}}
Since x and y are both positive
{{{x}}}{{{""=""}}}{{{y}}}

That is what we needed to prove.

Edwin</pre>