let a = arc distance of the sector
let r = radius of the circle
a + 2r = 8
a = (8-2r)
Find the portion of the circle in the sector (arc/circumference)
a/2r •r
Replace a with ( 8-2r)
(8-2r/2•square root • r)
Simply
(8-r/square root •r)
( 4-r) / square root R
Find the area of the sector
(4-r)/square root •r • square root •r ^2
Cancel pi*r
A = (4-r) * r
A = -r^2 + 4r
Max of this quadratic equation occurs at the axis of symmetry; x=-b/(2a)
R=-4/2•(-1)
R= 2 cm
Find the max area
A = -2^2 + 8(2)
A = -4+ 16
A = 12 sq cm max area
