Radius $r$ of the inner circle, radius $r_{1}$ of the four quarter circles. Diagonal of the unit square $\sqrt{2}=2r+2r_{1} \Rightarrow r_{1}=\frac{\sqrt{2}-2r}{2}$. Shaded area $=f(r)=1-\pi r^{2}-\pi(\frac{\sqrt{2}-2r}{2})^{2}$. Maximum if $f'(r)=0$, if $r=\frac{\sqrt{2}}{4}$, the maximum area is $1-\frac{\pi}{4}$. Minimum. a) if $r=\frac{1}{2}$, the minimum area is $1-\pi+\frac{\pi \sqrt{2}}{2}$. b) if $r_{1}=\frac{\sqrt{2}-2r}{2}=\frac{1}{2}$ or $r=\frac{\sqrt{2}-1}{2}$, the same minimum area.