Finding the point where an angle in a triangle is maximum 
As you can see in the image, I need to find the length L in terms of a and b given that y is the greatest angle it can possibly be for any value of L. I have tried to solve for y in terms of L and then differentiate it and set it equal to zero in order to find the maximum, however this creates an insanely complicated derivative. Thanks for your help.
 A: Consider the triangle KIO. Its area equals $bL$. It also equals $\frac{1}{2}\sin(y) |KI||KO|$. So maximizing $y$, which is equivalent in this case to maximizing $\sin(y)$, is equivalent to maximizing 
$\left(\frac{bL}{|KI||KO|}\right)^2=
\frac{b^2L^2}{(L^2+(a-b)^2)(L^2+(a+b)^2)}$, i.e. to minimizing $(L+\frac{(a+b)^2}{L})(L+\frac{(a-b)^2}{L})=L^2+((a+b)^2+(a-b)^2)+\frac{(a+b)^2(a-b)^2}{L^2}$, i.e. to minimizing $\ell + \frac{c}{\ell}$ in respect to $\ell$ after setting $L^2=\ell$ and $c=(a+b)^2(a-b)^2$. Now, $\ell+\frac{c}{\ell}$ is minimized for $\ell=\sqrt{c}$, and thus in the original problem $y$ is maximized for $L^2=\sqrt{(a+b)^2(a-b)^2}$ i.e. for $L=\sqrt{(a+b)(a-b)}$.
A: 
I have tried to solve for y in terms of L and then differentiate it and set it equal to zero in order to find the maximum, however this creates an insanely complicated derivative.

Let's try:
$$
y=\arctan{\frac{a+b}L}-\arctan{\frac{a-b}L}
$$
Hence:
$$
\frac{dy}{dL}=\frac{a-b}{L^2+(a-b)^2}-\frac{a+b}{L^2+(a+b)^2}=0\\
\quad\\
\implies 2b[(a+b)(a-b)-L^2]=0.
$$
Not so complicated.
