Answer :
The sector arc length is just a fraction of the circumference:
Laʀc = (2πR) • (θ ⁄ 2π) = R • θ ... where θ = central angle (of sector)
Ps = R + R + (R • θ) ... perimeter of sector = Ps
Ps = R • (2 + θ) ... Ps = 100 ft
100 = R • (2 + θ)
θ = (100 ⁄ R) – 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
Similarly, the sector area is just a fraction of the circle area:
As = (πR²) • (θ ⁄ 2π) ... area of sector = As
As = R² • θ ⁄ 2 ... substitute for θ
As = R² • [(100 ⁄ R) – 2 ] ⁄ 2
As = 50R – R² ... differentiate
As' = 50 – 2R ... set to zero
0 = 50 – 2R
R = 25 ft ... optimum radius
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
As = 50R – R² ... evaluate at: R = 25 ft
As = 50(25) – (25)²
As = 625 ft² ... maximum area
Note ... at any other R_value, the sector area is less.
" θ " can be determined using: θ = (100 ⁄ R) – 2
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Laʀc = (2πR) • (θ ⁄ 2π) = R • θ ... where θ = central angle (of sector)
Ps = R + R + (R • θ) ... perimeter of sector = Ps
Ps = R • (2 + θ) ... Ps = 100 ft
100 = R • (2 + θ)
θ = (100 ⁄ R) – 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
Similarly, the sector area is just a fraction of the circle area:
As = (πR²) • (θ ⁄ 2π) ... area of sector = As
As = R² • θ ⁄ 2 ... substitute for θ
As = R² • [(100 ⁄ R) – 2 ] ⁄ 2
As = 50R – R² ... differentiate
As' = 50 – 2R ... set to zero
0 = 50 – 2R
R = 25 ft ... optimum radius
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~...
As = 50R – R² ... evaluate at: R = 25 ft
As = 50(25) – (25)²
As = 625 ft² ... maximum area
Note ... at any other R_value, the sector area is less.
" θ " can be determined using: θ = (100 ⁄ R) – 2
-------------------------------------------------
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!