Question 288478
Alan is building a garden shaped like a rectangle with a semicircle attached to one short side.
 If he has 40 feet of fencing to go around it, what dimensions will give him the maximum area in the garden?
:
Short side = W
The half circumference of the semi circle = .5*pi*W
:
2Lengths + width + semicircle = 40 ft
2L + W + (.5*pi*W) = 40
2L + W + 1.57W = 40
Combine like terms
2L + 2.57W = 40
Simplify divide by 2
L + 1.285W = 20
L = (20-1.285W); for substitution
:
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 (20-1.285W)
A = W(20-1.285W) + .3925W^2
A = -1.285W^2 + .3925W^2 + 20W
A = -.8925W^2 + 20W
:
Find W by finding the axis of symmetry of this equation a=-.8925, b=20
W = {{{(-20)/(2*-.8925)}}}
W = {{{(-20)/(-1.785)}}}
W = +11.2 ft is the width for max area
then
L = 20 - 1.285(11.2)
L = 20 - 14 = 5.6 ft is the length
:
This presents a problem, the semi circle is supposed to be on the short side.
These calculations say the max area will be when the semi circle is on the long side.   
:
Check the perimeter
2(5.6) + 11.2 + 1.57(11.2) = 
11.2 = 11.2 + 17.6 = 40
:
I can't see how we could do it any other way
:
Rectangle: 5.6 by 11.2, semicircle circumference 17.6
:
Max area: (5.6*11.2) + (.5*pi*5.6^2) = 112 sq/ft