maximum area of semi-circle in square I'm struggling the with the following question:
Given is a square with length $a$. Now I want to find a semi-circle with the max. area. Looks like this:
http://tube.geogebra.org/material/show/id/222093
Now I want to prove, that the semi-circle is the best one (max. area) but have no idea, how to start.
 A: We solve the "dual" problem of finding the smallest square containing the semicircle $H:=\{(x,y)|x^2+y^2\leq 1, \ x\geq0\}$. Drawing supporting lines to $H$ having slopes $\phi$ and $\phi+{\pi\over2}$ with respect to the horizontal we see that the smallest square containing $H$ and having sides with these slopes has  side length
$$s(\phi)=\max\{1+\cos\phi,1+\sin\phi\}\qquad \bigl(0\leq\phi\leq{\pi\over2}\bigr)\ .$$
This is minimal when $\phi={\pi\over4}$, so that we obtain $s_{\min}=1+{\sqrt{2}\over2}$.
Returning to the original problem, where the side length $a$ of the square is given, we can say that the semicircle of maximal area inscribed in the square has its diameter parallel to a diagonal, and its radius $r_{\max}$ is given by
$$r_{\max}=\bigl(2-\sqrt{2}\bigr)\>a\ .$$
A: I'm not exactly sure how to prove that the construction in your worksheet has the max. area (I assume it does). However, to answer the question in your worksheet

What is the position of the center of the semicircle that produces the largest inscribed area in the square? What is this area?

For a given center $c=(n,n)$ you get a radius of $r=\sqrt2n$. You can maximize this by setting:
$$n+r=n+\sqrt2n=1$$
$$n=\sqrt2-1\approx0.41421 $$
and the area of the half-circle can then be computed knowing $r$.
