Given the base and vertical angle of a triangle show that its area is greatest when the triangle is isosceles.
For isosceles triangle (with base given 2x, and vertical angle let z):
Note: $y$ is got by $(180 - z)/2$
I calculated area to be: $$CD^2 * tan(y)$$ by using trigonometry and the fact that the altitude in isosceles triangle from vertex angle bisects the base.
I don't know how I can prove this area to be greatest.
Please help me. Thank you!