So I tried the good old Calculus 1 approach and turned this into an optimization problem. The equations got REALLY hairy, but it was okay since this was the graphing calculator section of the exam. I called the longer part of the horizontal diagonal $x_2$, the shorter part of the horizontal diagonal $x_1$, and half of the vertical diagonal $v$.
After getting $x_1$ in terms of $x_2$ and some tedious algebra, I got that $$2y=\sqrt{4 – \sqrt{x_2^2 – 21}} + \sqrt{25 – x_2^2}.$$ I then multiplied both sides of the equation by the total length of the horizontal diagonal, and graphed the right-hand side of the equation on my calculator. The logic was to find out where $(x_1 + x_2)2y$ would reach a maximum point, since the product of the diagonals of a kite equal twice the area of the kite.
After graphing that disaster I got a somewhat reasonable answer. I got the horizontal diagonal to be equal to roughly $4.5829$, and the vertical diagonal to be equal to $4$. However, I don't have an answer key, so I don't know if I am correct. Any feedback would be appreciated!
Best Answer
It is easier without a calculator. The area of the top triangle is $(1/2)(2)(5)\sin\theta$, where $\theta$ is the top angle. This is maximized when $\theta=\pi/2$. So the long diagonal has length $\sqrt{29}$. The other diagonal is now not hard to calculate, since the area of the kite is $10$.
Remark: In your notation, $x_1=\sqrt{x_2^2-21}$ and $y=\sqrt{25-x_2^2}$. So we are trying to maximize $$\sqrt{25-x_2^2}\left(\sqrt{x_2^2-21}+x_2\right).$$ From the description in the OP, it seems you may have used the wrong expression for $y$ in terms of $x_2$.