Math Competition Problem- Geometry/Calculus 
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!
 A: 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 $x_2$.
A: Using the long way by algebra.
As answered by André Nicolas,  we  try to maximize $$f(x_2)=\sqrt{25-x_2^2}\left(\sqrt{x_2^2-21}+x_2\right)$$ The derivative simplifies to $$f'(x_2)=-\frac{\left(\sqrt{x_2^2-21}+x_2\right) \left(x_2
   \left(\sqrt{x_2^2-21}+x_2\right)-25\right)}{\sqrt{(25-x_2^2) \left(x_2^2-21\right)}}$$ which can only cancel if  $$ x_2
   \left(\sqrt{x_2^2-21}+x_2\right)-25=0$$ that is to say $$\sqrt{x_2^2-21}=\frac {25}{x_2}-x_2$$ After squaring, the solution is $x_2=\frac{25}{\sqrt{29}}$ and, for this value, $f(x_2)=10$ 
A: Let $a, b, c$ be the length of the parts of the longer diagonal, and the half the shorter diagonal. Then: $S = ac + bc \leq \dfrac{a^2+c^2}{2}+\dfrac{b^2+c^2}{2}=\dfrac{2^2}{2}+\dfrac{5^2}{2}=\dfrac{29}{2}$
A: The upper (and/or) lower half of the kite is a triangle. Such a triangle with two given side lengths has maximal area when it is a right triangle. Now you can find the exact lengths of the diagonals with Pythagoras & Cie.
