A sector of a circle with radius r cm contains an angle of θ radians between the bounding radii. Given that the perimeter of the sector is 7cm, express θ in terms of r and show that the area is r/2(7-2r)cm square. Hence, find the maximum area of this sector as r varies.

Question

I have already done the first part of the question, i only don't know how to do the "hence...". Please help, Thanks!

Answer 1

arc length = a = rѲ
Then P = 2r + rѲ
7 = 2r + rѲ
Ѳ = (7-2r)/r

Let the area of the sector be A
by ratios
A/(pir^2) = Ѳ/(2pi)
A = (r^2)Ѳ/2
or
Ѳ = 2A/r^2

then 2A/r^2 = (7-2r)/r
solving for A after cross-multiplying and simplifying
A = (7r-2r^2)/2 or r/2(7-r) as required

second part:
A = (7/2)r - r^2 from above
dA/dr = 7/2 - 2r
= 0 for a max/min of A
2r - 7/2 = 0
r = 7/4
so Ѳ = (7 - 2(7/4))/(7/4)
Ѳ = 2 (how nice)

so max area = (r^2)(Ѳ)/2
= (49/16)(2)/(7/4)
= 7/2