What is the angle?

Question

Find the angle θ that maximizes the area of an isosceles triangle with legs of length L=2.

Answer 1

A = 1/2bh

sin(θ/2) = b/2 so b = 2sin(θ/2

cos(θ/2) = h/2 so h = 2cos(θ/2)

Putting these three together we get A(θ) = (1/2)(2sin(θ/2))(2cos(θ/2))

A(θ) = 2sin(θ/2)cos(θ/2)

Using the identity sin(2x) = 2sin(x)cos(x) we get:

A(θ) = sin(2•θ/2)

A(θ) = sin(θ)

To maximize the function, take the derivative and set it equal to zero:

A'(θ) = cos(θ)

cos(θ) = 0 when θ = π/2 (note that θ = 3π/2 does not work for a triangle)

So the maximum area occurs at θ = π/2