Label the vertices of the kite as above. Let x_1 = BD and x_2=AC, the lengths of the diagonals. Our goal is to find x_1 and x_2 such that the area of the kite A = (x_1x_2)/2 is maximal.
First we will write x_2 in terms of x_1, a, and b. Note that OD = (BD)/2 = x_1/2, AD = a, and CD = b.
Notice, before proceeding, that the triangle inequality applied to triangleABD gives x_1 < 2a, and applied to triangleBCD gives x_1<2b.
Applying the Pythagorean theorem to the right triangle AOD, we have AO^2+OD^2=AD^2
=> AO^2 = AD^2-OD^2
=> AO = sqrt(AD^2-OD^2) = sqrt(a^2-(x_1/2)^2)
Similarly, if we apply the Pythagorean theorem to the right triangle COD, we get CO^2 + OD^2 = CD^2
=> OC^2 = CD^2-OD^2
=> OC = sqrt(CD^2-OD^2) = sqrt(b^2-(x_1/2)^2)
Putting these together, we get x_2 as
x_2 = AC=AO+OC=sqrt(a^2-x_1^2/4)+sqrt(b^2-x_1^2/4)
(Note that the restrictions on x_1 mentioned above ensure that the arguments of the square roots are positive)
Next, we will find x_1 such that A=(x_1x_2)/2 is maximal.
Rewriting the area using the formula for x_2 we found above, we get
A = (x_1x_2)/2=1/2x_1(sqrt(a^2-x_1^2/4)+sqrt(b^2-x_1^2/4))
To maximize this, we will find its extrema by locating its critical values, points where its derivative is 0 or does not exist. The differentiation process is lengthy and is omitted to save space, but can be done using the product rule and chain rule. Doing so, together with some algebraic manipulation, we find (dA)/(dx_1) to be
((sqrt(4a^2-x_1^2)+sqrt(4b^2-x_1^2))(sqrt(4a^2-x_1^2)sqrt(4b^2-x_1^2)-x_1^2))/(4sqrt(4a^2-x_1^2)sqrt(4b^2-x_1^2))
which is equal to 0 if and only if one of
{(sqrt(4a^2-x_1^2)+sqrt(4b^2-x_1^2)=0), (sqrt(4a^2-x_1^2)sqrt(4b^2-x_1^2)-x_1^2=0):}
are true.
The first equation has no solutions given our restrictions, as both terms are positive and the sum of two positives cannot be 0. Solving the second:
sqrt(4a^2-x_1^2)sqrt(4b^2-x_1^2) - x_1^2 = 0
=> sqrt(4a^2-x_1^2)sqrt(4b^2-x_1^2) = x_1^2
=> (4a^2-x_1^2)(4b^2-x_1^2) = x_1^4
=> x_1^4-4a^2x_1^2 - 4b^2x_1^2 + 16a^2b^2 = x_1^4
=> 4(a^2+b^2)x_1^2 = 16a^2b^2
=> x_1^2 = (4a^2b^2)/(a^2+b^2)
As x_1 is a length, it must be positive, so we only need consider the positive square root of x_1^2.
:. x_1 = sqrt((4a^2b^2)/(a^2+b^2)) = (2ab)/sqrt(a^2+b^2).
And if we substitute our formula for x_1^2 into our formula for x_2, we get
x_2 = sqrt(a^2-(a^2b^2)/(a^2+b^2))+sqrt(b^2-(a^2b^2)/(a^2+b^2))
=sqrt((a^4+a^2b^2-a^2b^2)/(a^2+b^2))+sqrt((b^4+a^2b^2-a^2b^2)/(a^2+b^2))
=sqrt(a^4/(a^2+b^2))+sqrt(b^4/(a^2+b^2))
=(a^2+b^2)/sqrt(a^2+b^2)
=sqrt(a^2+b^2)
