Posted by Danny on Friday, June 19, 2009 at 6:17am.
arc length = a = rѲ
Then P = 2r + rѲ
7 = 2r + rѲ
Ѳ = (7-2r)/r
Let the area of the sector be A
A/(pir^2) = Ѳ/(2pi)
A = (r^2)Ѳ/2
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
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
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