Question 596257
<pre>
<font size = 4 color = "indigo"><b>
Let x represents half the length of the
rectangle, the length of rectangle = 2x

Let y represent the height of the rectangle. 

{{{drawing(400,196,-4,4,0,4,circle(0,0,2.1), line(-2,0,2,0),
rectangle(-1.29,0,1.29,1.6),locate(-.1,.18,o),
locate(-.7,.3,x), locate(.7,.3,x), locate(1.3,.9,y), locate(-.2,1.6,"2×") 

)}}}

Then the Area of the rectangle is

Area = length × width

   A = (2x)y

   A = 2xy

However we must now express y in terms of x and r.

Draw in a radius (which equals r) from the 
center of the semicircle to the upper right
corner of the rectangle:

{{{drawing(400,196,-4,4,0,4,circle(0,0,2.1), line(-2,0,2,0),
rectangle(-1.29,0,1.29,1.6),locate(-.1,.18,o),  line(0,0,1.3,1.6),
locate(-.7,.3,x), locate(.7,.3,x), locate(1.3,.9,y), locate(-.2,1.6,"2×"), 
locate(.4,1,"r")
)}}}

Use the Pythagorean theorem on the right triangle:

x² + y² = r²

     y² = r² - x²
           _______  
      y = <font face = "symbol">Ö</font>r² - x²

So substitute this for y in

      A = 2xy
             _______
      A = 2x<font face = "symbol">Ö</font>r² - x²

We could take the derivative in this form, but it'll be 
easier if we square both sides first to get rid of the 
square root:

     A² = 4x²(r² - x²)

     A² = 4r²x² - 4x<sup>4</sup>

Remember that r is a constant.


 2A{{{dA/(dx)}}} = 8r²x - 16x³

Solve for {{{dA/(dx)}}}

   {{{dA/(dx)}}} = {{{(8r^2x - 16x^3)/A}}}

We set that equal to 0 to find the extremum:

    {{{(8r^2x - 16x^3)/A}}} = 0

Multiply both sides by A

          8r²x - 16x³ = 0

         8x(r² - 2x²) = 0

     8x = 0;  r² - 2x² = 0

      x = 0;        r² = 2x²

Obviously the minimum area is when x = 0
                 only take positive square roots for equation on the right:
                          ___
                     r = <font face = "symbol">Ö</font>2x²
                           _
                     r = x<font face = "symbol">Ö</font>2

                     {{{r/sqrt(2)}}} = x

We find y from:

                         _______  
                    y = <font face = "symbol">Ö</font>r² - x²

We need to substitute x² = {{{r^2/2}}}

                    y = {{{sqrt(r^2 - r^2/2)}}}

                    y = {{{sqrt(2r^2/2 - r^2/2)}}}

                    y = {{{sqrt(r^2/2)}}}

                    y = {{{r/sqrt(2)}}}

So the largest rectangle has base 2x = {{{2r/sqrt(2)}}}

and height y = {{{r/sqrt(2)}}}

To find the maximum area, substitute in

                    A = (2x)y

                    A = {{{2r/sqrt(2)}}}{{{""*""}}}{{{r/sqrt(2)}}}
                   
                    A = {{{2r^2/2}}}

                    A = r²


Edwin</pre>