A square and two rectangles in a round hole

Without loss of generality let the circle have radius $1$ , making $C = \pi$ .

Now the diagonal of the square section is a diameter of the circle $x^{2} + y^{2} = 1$ and hence has length $2$ . The sides of the square thus have length $\sqrt{2}$ , making its area $2$ .

Now let the center of the circle (and square) coincide with the origin of a superimposed $xy$ -grid, with the square symmetric about both the $x$ and $y$ axes. The top side of the square then lies on the line $y = \frac{1}{\sqrt{2}}$ . We can assume that the red rectangle at the top of the diagram has maximum area when it is symmetric about the $y$ -axis. Letting its right side have horizontal component $x \gt 0$ , its area $A$ will be

$A = 2x*(\sqrt{1 - x^{2}} - \frac{1}{\sqrt{2}})$ .

To find any critical points we then set $\frac{dA}{dx} = 0$ :

$\frac{dA}{dx} = 2*(\sqrt{1 - x^{2}} - \frac{1}{\sqrt{2}}) + 2x*(\frac{-x}{\sqrt{1 - x^{2}}}) = 0$

$\Longrightarrow \sqrt{1 - x^{2}} - \frac{x^{2}}{\sqrt{1 - x^{2}}} = \frac{1}{\sqrt{2}}$

$\Longrightarrow \frac{1 - 2x^{2}}{\sqrt{1 - x^{2}}} = \frac{1}{\sqrt{2}}$

$\Longrightarrow 2*(1 - 2x^{2})^{2} = 1 - x^{2} \Longrightarrow 8x^{4} - 7x^{2} + 1 = 0$ ,

which is quadratic in $x^{2}$ with solutions

$x^{2} = \frac{7 \pm \sqrt{17}}{16} \Longrightarrow x = \frac{\sqrt{7 \pm \sqrt{17}}}{4}$ .

Now note that the coordinates of the top right vertex of the square is $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ , and thus we must have $x \le \frac{1}{\sqrt{2}}$ . Only one of our potential solutions meets this requirement, namely

$x = \frac{\sqrt{7 - \sqrt{17}}}{4}$ .

This gives us the value $A = (\frac{\sqrt{7 - \sqrt{17}}}{8})*(\sqrt{9 + \sqrt{17}} - 2\sqrt{2})$ .

Given that the minimum possible value for the area of the red rectangle, (ranging over the possible values for $x$ ), is zero, we can surmise that the value for $A$ found above is a maximum. By symmetry, the red rectangle on the right side of the square has the same maximum area value. The maximum possible ratio of the cut-out area to the area of the circle is then

$\frac{C}{S} = \frac{2 + 2A}{\pi} = 0.7438106....$ , making $\lceil 1000*\frac{C}{S} \rceil = \boxed{744}$ .

We'll work assuming that the circle's radius is 1 just to keep our solution pretty.

We begin with making a circle with the equation $x^2+y^2=1$

The inscribed square has a height of $\sqrt2$ , thus the distance from the center to the top is $\frac{\sqrt{2}}{2}$ .

To make the top side even with the x axis, we must make a vertical shift of $\frac{1}{\sqrt{2}}$ units down and our graph becomes

$x^2+(y+\frac{1}{\sqrt2})^2=1$

$\Rightarrow y=\sqrt{1-x^2}-\frac{1}{\sqrt{2}}$

Now, looking at the rectangle, we see that the base is along x=0, thus the base length will be $2x$ (x is the coordinate of the base along the x-axis).

The height $y_i$ will be corresponding to $x_i$ . Thus using the formula area=base*height, we get $A=(2x)(\sqrt{1-x^2}-\frac{1}{\sqrt{2}})$

To maximize this, we take the derivative $\dfrac{dA}{dx}$ and set it equal to 0

$\frac{dA}{dx}=(2x)(\sqrt{1-x^2}-\frac{1}{\sqrt{2}})$

We differentiate in parts

$0=(2)(\sqrt{1-x^2}-\frac{1}{\sqrt{2}})+(2x)(\dfrac{-x}{\sqrt{1-x^2}})$

Multiplying out the denominator on both sides and some algebraic manipulation yields $8x^4-7x^2+1=0$

Using the quadratic then square rooting, we get $x=\pm\dfrac{\sqrt{7-\sqrt{17}}}{4}$

"Wolfing around" a little with our value of x, plugging it into the equation $A=(2x)(\sqrt{1-x^2}-\frac{1}{\sqrt{2}})$ .

We get the area of one rectangle to be $\frac{1}{8}\left(-2 \sqrt{14-2 \sqrt{17}}+\sqrt{46-2 \sqrt{17}}\right)$

Multiplying this by 2 (there are two rectangles) and adding the area of the cut out square (2 sq units) and finally dividing everything by the area of the circle ( $\pi$ sq units we get $\frac{C}{S}=\dfrac{\frac{1}{4} \left(-2 \sqrt{14-2\sqrt{17}}+\sqrt{46-2 \sqrt{17}}\right)+2}{\pi}\approx 0.7438$

Therefore $\boxed{\lceil \left(1000\frac{C}{S} \right) \rceil=744}$