Find the dimensions of a rectangle of maximum size that will fit into a half circle of radius 27

Make the rectangle sit on the diameter.

Let its base be 2x and its height be y
then x^2 + y^2 = 27^2
y = √(729 - x^2)

area = 2xy
area^2 = 4x^2y^2
= 4x^2(729-x^2)
= 2916x^2 - 4x^4
d(area^2)/dx = 5832x - 16x^3
= 0 for a max of area

(if area is the largest of all areas, then area^2 will be the largest of the squares of the area)

5832x - 16x^3 = 0
8x( 729 - 2x^2) = 0
8x = 0 ---> x = 0 , would give no rectangle at all (a min)
or
729 = 2x^2
x^2 = 729/2
x = 27/√2 or 27√2/2

so 2x = 27√2
and from x^2 + y^2 = 729
729/2 + y^2 = 729
y^2 = 729/2
y = 27/√2 = 27√2/2

So the rectangle is 27√2 by 27√2/2