Question 1151004: A community beach wants to rope off a swimming area of 400m^2. Should they create a semi circular or rectangular area if they want to minimize the rope needed?(I believe semi-circular, but can't explain!)
Found 2 solutions by rothauserc, jim_thompson5910:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Area of the semi-circle = (pi * r^2)/2
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(pi * r^2)/2 = 400
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r^2 = 800/pi
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r = 15.9577 m
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Circumference of the semi-circle = (pi * diameter)/2 = pi * 15.9577 = 50.1326 m
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Area of a rectangle = length * width
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Note all squares are rectangles but all rectangles are not squares
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A square will yield the minimum perimeter, in this case we only need to consider three sides since the beach will be the other side :-)
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Area of the square = s^2 = 400, therefore the side is 20 m
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The perimeter required is 60 m, therefore the circumference(50.13226 m) of semi-circle is < the perimeter required for the square(minus one side for the beach) 60 m
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Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
You have the correct answer. Below is the explanation why.
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We are given an area of 400 square meters to work with. Let's see how much rope we need if we want to make a semi-circle with this area.
Use the area of a semi-circle formula to find what the radius R must be so that we enclose an area of A = 400 square meters.
A = 400
A = (pi*R^2)/2 ... general area of semi circle
(pi*R^2)/2 = 400
pi*R^2 = 2*400
pi*R^2 = 800
R^2 = 800/pi
R = sqrt(800/pi)
A negative R value is not possible, so we don't have to worry about the plus/minus.
Now compute the distance along the curved edge of the semi-circle (aka the distance along half of a circle)
P = distance around curved edge of semi-circle
P = (full circle circumference)/2
P = (2*pi*R)/2
P = pi*R
P = pi*sqrt(800/pi) ........ substitution
P = sqrt(pi^2)*sqrt(800/pi)
P = sqrt(pi^2*800/pi)
P = sqrt(800pi)
P = sqrt(400*2pi)
P = sqrt(400)*sqrt(2pi)
P = 20*sqrt(2pi)
P = 50.132565 ........ use your calculator here
So we need roughly 50.132565 meters of rope if we want to form a semi-circular enclosure of area 400 square meters.
We will use this value for comparison purposes later. Let M = 50.132565
We don't need rope along the straight line portion as the beach itself has already taken care of this part.
The beach forms a natural wall or barrier. This diagram shows what I mean
The diagram is not to scale. The dashed red line is where the ocean meets the beach. Everything above the dashed line is the ocean. Everything below the dashed line is the beach. Rope is not needed along the dashed red line.
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Now let's consider the rectangular enclosure.
L = length of rectangle = side length parallel along beach
W = width of rectangle = side perpendicular to beach
A = area of rectangle
A = L*W
LW = A
LW = 400
W = 400/L
Refer to the diagram above. We have 1 side of L and 2 sides of W. Like with the semi-circle, we don't need rope along the bottom side of the rectangle.
The three sides add to: W+W+L = 2W+L
Let P be the total distance
P = 2W+L
P = 2(400/L)+L ... plug in W = 400/L
P = 800/L+L
P = 800/L+(L^2)/L
P = (800+L^2)/L
Apply the derivative to find when its perimeter P is at its minimum (see note 1 below at the bottom of the page)
P = (800+L^2)/L
P = n/d
n = 800+L^2 ... is the numerator function
d = L ... is the denominator function
n' = 2L ... derivative, with respect to L, of the numerator function
d' = 1 ... derivative, with respect to L, of the denominator function
P = n/d
P' = (n'*d - n*d')/(d^2) .... quotient rule
P' = (2L*L - (800+L^2)*1)/(L^2) ... substitution
P' = (2L^2 - (800+L^2))/(L^2)
P' = (2L^2 - 800 - L^2)/(L^2)
P' = (L^2 - 800)/(L^2)
Solve P' = 0 for L
P' = (L^2 - 800)/(L^2)
0 = (L^2 - 800)/(L^2)
0 = L^2 - 800
L^2 = 800
L = sqrt(800)
L = sqrt(400*2)
L = sqrt(400)*sqrt(2)
L = 20*sqrt(2)
W = 400/L
W = 400/(20*sqrt(2))
W = 20/(sqrt(2))
W = (20*sqrt(2))/(sqrt(2)*sqrt(2))
W = (20*sqrt(2))/(2)
W = 10*sqrt(2)
This rectangle has dimensions
L = 20*sqrt(2) as the length
W = 10*sqrt(2) as the width
Something of interest: The length is twice as long as the width, so L = 2*W. This happens with any area value and not just with A = 400.
Because  , this means we do not have a square forming (see note 2 below at the bottom of the page). A square is only possible if all four side lengths of the figure are the same.
Let's compute the value of P, which was the distance around the sides of the rectangle in the ocean water.
P = 2W+L
P = 2*10*sqrt(2)+20*sqrt(2)
P = 20*sqrt(2)+20*sqrt(2)
P = 40*sqrt(2)
P = 56.568542
We need roughly 56.568542 meters of rope if we want to form a rectangular enclosure that has area 400 m^2 and the amount of rope needed has been minimized as much as possible.
We will use this value for comparison purposes later. Let N = 56.568542
In summary, we found
M = length of rope needed for the semi-circular enclosure = 50.132565 meters
N = length of rope needed for the rectangular enclosure = 56.568542 meters
Both values are approximate.
Since M < N, this means the semi-circular enclosure requires less rope. By design, both enclosures have the same area of 400 m^2.
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Notes
Note 1: if you aren't familiar with calculus and derivatives, then you can use a calculator to find the lowest point of
f(x) = (800+x^2)/x
which is a slight variation of
P = (800+L^2)/L
but I replaced L with x and P with f(x). Plotting that graph and using your calculator's "minimize" feature, you should be able to locate the (x,y) point that provides the smallest value of P. Keep in mind that x = L and y = P in this case. Also keep in mind that x > 0 and y > 0. Negative lengths and perimeters are not possible. Therefore you should only focus on the first quadrant.
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Note 2: The tutor @rothauserc has the right general idea/steps and the correct final answer; however, they made an error in stating that a square is what minimizes the amount of rope needed for the rectangular enclosure. If we had all four sides of a rectangle to work with, then a square would minimize the perimeter.
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