Find the angle theta that maximizes the area of an isosceles triangle whose legs have length l. The angle is the top angle if the left and right sides are l coming to a point with the bottom leg horizontal.
The Attempt at a Solution
I broke the triangle up into two halves to use right angle trig and eventually got the area to equal A=l^2 * sin(theta/2)*cos(theta/2). When I took the derivative though I realized that I would have too many variables. I think theres a way to solve for l in terms of theta or theta in terms of l but I'm not sure how to do it can anyone point me in the right direction.