Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 6 The area of

the isosceles triangle is A =

Let Q be the apex angle.
Draw the altitude of the triangle (bisecting the apex angle) through the center of the circle.
This altitude has length 6 + h, where h = 6 cos(Q)
Base of triangle has length: 12 sin(Q)
So the area is: A = (1/2)(6 + 6 cos(Q) )(12 sin(Q) = 36 sin(Q) + 18 sin(2Q)
dA/dQ = 36 cos(Q) + 36 cos(2Q)
Set this expression to zero to find the optimal triangle.
cos(Q) + 2cos^(Q) - 1 = 0
(2 cos(Q) - 1)(cos (Q) + 1 ) = 0 ; which gives: cos(Q)= 1/2 OR cos(Q) = -1
Rejecting the second solution, we have Q = pi/3 , so the triangle is equilateral.
The maximal area is: 36(3/2)(sqrt(3)/2) = 27 sqrt(3)