Finding maximal-sized rectangle inside circular segment

Question

Assume we have a circle with radius $r$ and its segment with height $v<r$. I want to fit in a rectangle whose area is the biggest. My approach was as follows:

Because $r>v$, we can describe a half circle with equation $f(x)=\sqrt{r^2-x^2}-(r-v)$. It is obvious that one of the vertices lies on this circle. Any point at the given circle has coords $[x,f(x)]$ We can then compute the area of such rectangle as follows: $A=2xf(x)$. To find the max, we define a new function $g(x)=2xf(x)$ for the derivative, we obtain: $$g'(x)=2\frac{-2x^2+r^2-(r-v)\sqrt{r^2-x^2}}{\sqrt{r^2-x^2}}$$ setting it equal to zero gives: $$-2x^2+r^2-(r-v)\sqrt{r^2-x^2}=0$$ Here I am stuck. I tried to square root both sides (moving the square root to the right hand side) but that leaves me with really messy expression with $r,v$ in degrees $>2$. If i simplify, i get following equation in $x$: $$4x^4+(-3r^2-2rv+v^2)x^2+2r^3v-v^2r^2$$ Using Discriminant tells me absolutely nothing about which of the two roots should i take (whether + or - $D$: $$D=(-3r^2-2rv+v^2)^2-16(2r^3v-v^2r^2)$$ then: $$x^2_{1,2}=\frac{3r^2+2rv-v^2\pm\sqrt{D}}{8}$$ What should i do next? Or would anyone give more elegant approach?

This picture should sum up, what I am dealing with https://i.stack.imgur.com/spcuI.jpg Let me explain what i added here. Green is the equation of the circle, Red is the function that we are finding the maximum of Blue is its derivative h(x) - orange is the polynomial i posted here in my steps q(x) - purple is the polynomial after substituting $x^2\to x$

Answer 1

Your equation for a vanishing $g'(x)$ can be written as $$ -2x^2+r^2=(r-v)\sqrt{r^2-x^2}. $$ The right hand side is positive, while the left hand side is positive only if $2x^2<r^2$.

Setting $w=r-v$ (with $-r\le w\le r$) your solution can be written as: $$ x^2=\frac{4r^2-w^2\pm\sqrt{w^4+8r^2w^2}}{8}, $$ that is: $$ x^2=\frac{4r^2-w^2\pm|w|\sqrt{w^2+8r^2}}{8}. $$ We can check that for the largest solution $2x^2>r^2$: $$ 2x^2=\frac{4r^2-w^2+|w|\sqrt{w^2+8r^2}}{4}= r^2+\frac{|w|\sqrt{w^2+8r^2}-w^2}{4}\ge r^2, $$ where the equal sign holds only for $w=0$, in which case both solutions are the same. This cannot then be a valid solution and must be discarded.

Answer 2

We should choose the $x$ value with $-\sqrt{D}$. (This is because the other choice of $x$ value would not yield $g'(x)=0$. A straightforward way to check this is to calculate $D$, $x$, $g'(x)$ for specific values of the parameters $r$ and $v$ given in your assignment.)

For example, plugging in $r=10, \ v=1$, we find $D=9^2\cdot881$, and $$ \max_{x>0}\{x (\sqrt{100 - x^2} - 10 + 1) \} = 190 \sqrt{38\over 235439 + 7929 \sqrt{881}} $$ at $$ x = \sqrt{ {319 - 9 \sqrt{881}\over8}}, $$

while $$ \sqrt{ {319 + 9 \sqrt{881}\over8}} \approx 8.56 $$ and $$ g' \left( \sqrt{ {319 + 9 \sqrt{881}\over8}} \right) \ne 0. $$