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March 2, 2015

March 2, 2015

Posted by **Danny** on Friday, June 19, 2009 at 6:17am.

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**, Friday, June 19, 2009 at 1:46pmarc 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

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