Common area of two squares Two squares are inscribed in a  circle of diameter $\sqrt2$ units.prove that common region of squares is at least 2$(\sqrt2-1)$
This question is in my calculus book but  seems to be algebraic.but frankly speaking i have no idea how to proceed .

Here is one way.

I drew two squares inscribed in the circle centered at the origin with radius $\frac{\sqrt 2}2$. The equation of the line containing the upper side of the red square, which is parallel to the axes, is $y=\frac 12$. The blue square is at an angle of $\alpha$ to the red square. The overlap area is eight times the area of the green triangle. The altitude of the green triangle is $\frac 12$, so to minimize its area we just minimize the length of the triangle's side.
Therefore, use geometry and/or trigonometry to find the length of that side of the triangle, and use trigonometry and/or calculus to find the minimum length of that side, given $0\le\alpha\le \frac{\pi}4$. You will find that the maximum is at $\alpha=0$ and minimum is at $\alpha=\frac{\pi}4$. Calculate the triangle's area at $\alpha=\frac{\pi}4$ and multiply by $8$, and you have your desired minimum overlap area.
I'll give you one more hint: the length of that triangle side is
$$\frac 1{1+\sin\alpha + \cos\alpha}$$
but there are also other ways to express it.
Ask if you need more help, but show some of your own work and state where you are stuck.
