Let the short side be W
The half circumference of the semi-circle = .5 * pi *W
2Lengths + width + semicircle = 20 ft
2L + W + (.5*pi*W) = 20
2L + W + 1.57W = 20
Combine like terms
2L + 2.57W = 20
Simplify divide by 2
L + 1.285W = 10
L = (10 - 1.285W); for substitution:
Get the area equation:
Rectangle area + semicircle area
A = L * W + (.5*pi*(.5W) ^2)
A = LW + (1.57*.25W^2)
A = LW + .3925W^2
Replace L with (10-1.285W)
A = W(10-1.285W) + .3925W^2
A = -1.285W^2 + .3925W^2 + 10W
A = -.8925W^2 + 10W
Find W by finding the axis of symmetry of this equation a=-.8925, b=10
W = -10 / 2 * (-.8925)
W = -10 / -1.785
W = 5.60 is the width for max area
then we solve for the length
L = 10 - 1.285(5.60)
L = 10 – 7.20 = 2.8 ft is the length
Then, Check the perimeter
2(2.8) + 5.60 + 1.57(5.60) = 20ft
5.6 + 5.6 + 8.736 = 20ft
Rectangle: 2.8 by 5.60 has semicircle circumference 8.736 has a maximum area: (2.8*5.60)
+ (.5 * pi * 2.8 ^ 2)
=15.68 + 12.31
=28 sq/ft
