Given a halved circle with radius $R$ and four rectangles inside it such that one of the side touches the middle line of the circle. If the area of the four rectangles are maximum, what is the area between the rectangles and the halved circle?
I'm sorry for my bad English so I try to illustrate it such follows.
We need to find the area of grey colored area. I really don't have any idea what to do. If there is only one big rectangle in that semi-circle, I think I know what to do. But what if the rectangles are more than one and the length of each rectangles sides can be vary?
The answer ist going to be symmetric so we will focus on how to maximize the area of two rectangles inside an eighth of a circle. Consider a point $P(x,y)$ inside this sector at the bottom corner of the green rectangle from the figure above. We construct the edges of the two bottom rectangles like this:
Call the width of the bottom rectangle $a$ and the sum of their heights $b$. Then the area of the blue hexalateral is:
$$A=bx+(a-x)y-\frac{b^2}{2}=x\sqrt{r^2-x^2}+y\left(\sqrt{r^2-y^2}-x\right)-\frac{r^2-x^2}{2}$$
In case you aren't familiar with multivariable calculus, partial differentiation works the same as ordinary while treating any other variables as constants. To find the maximum of A, we need the zeros of its partial derivatives:
$$\frac{\partial A}{\partial x}=\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}+x-y=0 \qquad y=\frac{r^2-2y^2}{\sqrt{r^2-x^2}}+x$$ $$\frac{\partial A}{\partial y}=\sqrt{r^2-y^2}-\frac{y^2}{\sqrt{r^2-y^2}}-x=0 \qquad y=\sqrt{\frac{r^2}{2}-\frac{x}{8}\left(x+\sqrt{8r^2+x^2}\right)}$$
Using a numerical method, we obtain $x \approx 0.795r$ and $y \approx 0.359r$, so $A \approx 0.3479r^2$.
So for a semicircle the minimal area not covered by four rectangles is $4r^2\left(\frac{\pi}{8}-A\right) \approx 0.179r^2$.