Please refer to the attached figure.
If the circle has radius [tex] R [/tex], the semicircle has equation [tex] y = \sqrt{R^2-x^2} [/tex], defined for [tex] -R < x < R [/tex].
The coordinates of the points are:
[tex] C' = (x,0) [/tex]
[tex] C = (-x,0) [/tex]
[tex] D = (-x,\sqrt{R^2-x^2}) [/tex]
[tex] E = (x,\sqrt{R^2-x^2}) [/tex]
This implies that
[tex] \overline{CC'} = 2x,\qquad \overline{CD} = \sqrt{R^2-x^2} [/tex]
So, the area of the rectangle is the product of the width and the height. We want to find the maximum of
[tex] f(x) = 2x\sqrt{R^2-x^2} [/tex]
To do so, let's start by computing its derivative:
[tex] f'(x) = 2\sqrt{R^2-x^2} + 2x \frac{-2x}{2\sqrt{R^2-x^2}} = 2\sqrt{R^2-x^2} - \frac{4x^2}{\sqrt{R^2-x^2}} = \frac{2(R^2-2x^2)}{\sqrt{R^2-x^2}} [/tex]
The derivative equals zero if the numerator equals zero:
[tex] 2(R^2-2x^2) = 0 \iff R^2-2x^2 = 0 \iff x^2 = \frac{R^2}{2} \iff x = \frac{R}{\sqrt{2}} [/tex]
So, we have
[tex] \overline{CC'} = 2\frac{R}{\sqrt{2}} = R\sqrt{2},\qquad \overline{CD} = \sqrt{R^2-\frac{R^2}{2}} = \sqrt{\frac{R^2}{2}} = \frac{R}{\sqrt{2}} [/tex]
