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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.

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    "y is the greatest for any possible value..." **of what** ? What is l...or 1 ...or that thing? – 2017-02-27
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    l is the distance from A to K, so in terms of a and b (which are the lengths at the bottom) what value of l gives the greatest angle y – 2017-02-27
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    Do you have the solution available? – 2017-02-27
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    I'm pretty sure it's root(a^2-b^2) – 2017-02-27
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    I still can't get it: **what is given**? The lower leg's length $\;a+b\;$ and the hypotenuse's? How is $\;y\;$ decided/formed? This is pretty confusing... – 2017-02-27
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    The question seems fairly clear to me: a and b are given, AI=a-b and AO=a+b, L is the variable you can adjust to maximize y. Note that "at a glance" $y\to 0$ if either $L\to\infty$, or if $L\to 0$. – 2017-02-27

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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 $L=\sqrt{(a+b)(a-b)}$.