The perimeter of a sector is constant. If its area is to be maximum, what is the central angle of the sector?

Level: Senior

Let t be the central angle of the sector. So the perimeter and the area of the sector are C = (t+2)r and A = tr^2/2. By the AM-GM inequality,
P^2
= (tr/2+tr/2+r+r)^2
>= {4[(tr/2)(tr/2)*r*r]^(1/4)}^2
= 8tr^2
= 16A^2,
with equality iff
tr/2 = r,
t = 2.
Thus A has a maximum value of P/4 and the corresponding central angle of the sector is 2 radians.