<PRE><B>Question 1128652</B>
&lt;br&gt;
Logical reasoning says the rectangle with the greatest area inscribed in a circle is a square.  The diameter of the circle is then the diagonal of the square.&lt;br&gt;
With a diagonal of length 20, the side of the square is 10*sqrt(2); the area is side squared = 200.&lt;br&gt;
To verify that answer, consider the circle with center at the origin, so the equation is x^2+y^2 = 100.  A point on the circle has coordinates {{{x}}} and {{{sqrt(100-x^2))}}}.&lt;br&gt;
The dimensions of the rectangle determined by that point are {{{2x}}} and {{{2*sqrt(100-x^2)}}}; the area is {{{4x(sqrt(100-x^2))}}}.&lt;br&gt;
The maximum area is when the derivative of the area function is 0.&lt;br&gt;
To simplify the process of finding where the derivative is 0, move all the variables inside the radical:&lt;br&gt;
{{{A = 4*sqrt(100x^2-x^4)}}}&lt;br&gt;
Set the derivative equal to 0 and solve:&lt;br&gt;
{{{(4(200x-4x^3))/(2*sqrt(100x^2-x^4)) = 0}}}
{{{(16x(50-x^2))/(2*sqrt(100x^2-x^4)) = 0}}}
{{{50-x^2 = 0}}}
{{{x^2 = 50}}}
{{{x = sqrt(50)}}}&lt;br&gt;
The dimensions of the rectangle with greatest area are&lt;br&gt;
{{{2x = 2*sqrt(50)}}}  and  {{{2(sqrt(100-x^2)) = 2*sqrt(50)}}}&lt;br&gt;
The area is&lt;br&gt;
{{{(2*sqrt(50))(2*sqrt(50)) = 200}}}</PRE>