Posted by Danny on .
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(72r)cm square. Hence, find the maximum area of this sector as r varies.
I have already done the first part of the question, i only don't know how to do the "hence...". Please help, Thanks!

How do i do this maths question?? 
Reiny,
arc length = a = rѲ
Then P = 2r + rѲ
7 = 2r + rѲ
Ѳ = (72r)/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 = (72r)/r
solving for A after crossmultiplying and simplifying
A = (7r2r^2)/2 or r/2(7r) 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