SOLUTION: 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 maxi
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-> SOLUTION: 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 maxi
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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? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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 =
W =
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
