place your sketch on the x-y grid with the centre at (0,0)
let the radius of the semicircle be r, where r is a constant
so the base of the rectangle is 2x, let its height be y
then x^2 + y^2 = r^2
y =(r^2 - x^2)^(1/2)
area = 2xy
d(area)/dx = (2x)(1/2)(r^2 - x^2)^(-1/2) (-2x) + 2(r^2-x^2)^(1/2)
= 0 for a max/min of area
2(r^2 - x^2)^(1/2) = 2x^2/(r^2-x^2)^(1/2)
x^2 = r^2 - x^2
2x^2 = r^2
x^2 = r^2/2
x = r/√2 = (√2/2)r
then largest area
= 2x√(r^2 - x^2)
= (4√2/2)r (r^2 - r^2/2^(1/2)
= 2√2r(r/√2) = 2r^2
or, in a real simple way
suppose we look at the whole circle, the largest "rectangle we can fit inside the circle is a square, where x = y
then the sides of the square are 2x and 2y,
and the area is 4xy, but x=y and x^2 + x^2 = r^2 ---> x^2 = r^2/2
so the largest area = 4x^2
= 4(r^2/2) = 2r^2
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