SOLUTION: FInd the area of the largest rectangle that can be inscribed in a semicircle of fadius r.

Let x represents half the length of the
rectangle, the length of rectangle = 2x

Let y represent the height of the rectangle. 



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:



Use the Pythagorean theorem on the right triangle:

x² + y² = r²

     y² = r² - x²
           _______  
      y = Ör² - x²

So substitute this for y in

      A = 2xy
             _______
      A = 2xÖr² - 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 - 4x4

Remember that r is a constant.


 2A = 8r²x - 16x³

Solve for 

    = 

We set that equal to 0 to find the extremum:

     = 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 = Ö2x²
                           _
                     r = xÖ2

                      = x

We find y from:

                         _______  
                    y = Ör² - x²

We need to substitute x² = 

                    y = 

                    y = 

                    y = 

                    y = 

So the largest rectangle has base 2x = 

and height y = 

To find the maximum area, substitute in

                    A = (2x)y

                    A = 
                   
                    A = 

                    A = r²


Edwin