The Rancher Barney Problem

Transcription

The Rancher Barney Problem
Alex Bilzerian, Natasha Honcharik, Ashwin Panda, Juliana
West
The Problem
Rancher Barney has a prize bull and some cows on his ranch. He has a large area for pasture
that includes a stream running along one edge. He must divide the pasture into two regions, one region
large enough for the cows and the other, smaller region to hold the bull.
bull’s pasture must be at least 1,000 square meters for grazing and the cow pasture must be at
least 10,000 square meters to provide grazing for the cows. The shape of the pasture is basically a
rectangle 120 meters by 150 meters. The river runs all the way along the 120 meter side. Fencing
costs $5/meter and each fence post costs $10. Any straight edge of fence requires a post every 20
meters and any curved length of fence requires a post every 10 meters.
Prepare a proposal from your team to rancher Barney that will minimize his total cost of fencing.
Ashwin Panda' s Method
HSimilar to Juliana' s ApproachL
Approach #1 - Make the bull pasture the smallest possible square along the river, and use the
remaining edge across the river to serve as the base for the cow pasture, which would be a rectangle.
Assumptions: This method assumes that both the cows and the bull must be completely enclosed. It
also assumes that there does have to be a post at each corner or end of the fence and that there does
not need to be fencing along the river.
Printed by Wolfram Mathematica Student Edition
page 1 of 13
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Group B
Alex Bilzerian - Club
Area1 = 1000.0;
x = Sqrt@1000D;
Area2 = 10 000.0;
y = Area2 H120.0 - xL;
Perimeter = 2 x + 2 y + H120.0 - xL;
FencePrice = 5.00;
FenceCost = FencePrice * Perimeter;
Posts = Round@HPerimeter 20.0L + 7.00D;
PostCost = 10.0 * Posts;
TotalCost = FenceCost + PostCost
2149.63
Natasha’s Method
Approach #2The object here was to make a function that gives the portion of the perimeter of the two enclosures that
would require fencing in terms of x, the width of the cow pen along the river. Then this function can be
minimized to find the value of the distance x (and thus all the other enclosure dimentions) such that the
length of fence needed is as small as possible for rectangular pens.
Assumptions: This method assumes that both the cows and the bull must be completely enclosed. It is
assumed that there does not need to be fencing along the river. Like the previous approach this
assumes that there must be a post at every corner or end of fence. But it also minimizes the number of
posts needed by arranging them in increments of the maximum 20 meters and adds a post at the next
corner only if it is necessary.
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B
Assumptions: This method assumes that bothGroup
the cows
and the bull must be completely enclosed.
It is
assumed that there does not need to be fencing along the river. Like the previous approach this
assumes that there must be a post at every corner or end of fence. But it also minimizes the number of
posts needed by arranging them in increments of the maximum 20 meters and adds a post at the next
corner only if it is necessary.
verticies1 = 880, 0<, 82, 0<, 82, - 3<, 80, - 3<, 80, 0<<;
verticies2 = 882, 0<, 83, 0<, 83, - 1<, 82, - 1<, 82, 0<<;
verticies3 = 880, 0<, 83, 0<, 83, - 4<, 80, - 4<, 80, 0<<;
pasture = Graphics@88GrayLevel@.8D, Polygon@verticies1D<,
8GrayLevel@.8D, Polygon@verticies2D<, 8Line@verticies3D<<D;
text1
text2
text3
text4
text5
=
=
=
=
=
Graphics@Text@"120", 82, .1<DD;
Graphics@Text@"150", 8- .25, - 2<DD;
Graphics@Text@"A=10000", 81, - 2<DD;
Graphics@Text@"A=1000", 82.5, - .5<DD;
Graphics@Text@"x", 81, - .1<DD;
Show@pasture, text1, text2, text3, text4, text5, riverD
120
x
A=1000
150
A=10000
perimeter = x + H120 - xL + H2 * 10 000 xL + 1000 H120 - xL
1000
120 +
20 000
+
120 - x
x
1000
function = PlotBy = 120 +
point = 8898.1, 370<<;
20 000
+
120 - x
x
, 8x, 0, 120<, PlotRange -> 80, 1000<F;
minimum = ListPlot@point, PlotStyle ® PointSize@.03DD;
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Group B
Show@function, minimumD
1000
800
600
400
200
0
20
40
60
80
100
120
x = 98.1
1000
perimeter = 120 +
369.536
120 - H98.1L
20 000
+
H98.1L
fencecost = 5 perimeter
1847.68
verticies4 = 880, 0<, 898.1, 0<, 898.1, - 101.96<, 80, - 101.96<<;
verticies5 = 8898.1, 0<, 8120, 0<, 8120, - 45.66<, 898.1, - 45.66<<;
pasture2 = Graphics@
88GrayLevel@.8D, Polygon@verticies4D<, 8GrayLevel@.8D, Polygon@verticies5D<<D;
points = 880, 0<, 80, - 20<, 80, - 40<, 80, - 60<, 80, - 80<,
80, - 100<, 80, - 101.96<, 898.1, 0<, 898.1, - 20<, 898.1, - 40<,
898.1, - 45.66<, 8120, 0<, 8120, - 20<, 8120, - 40<, 8120, - 45.66<,
8110, - 45.66<, 898.1, - 65<, 898.1, - 85<, 898.1, - 101.96<, 820, - 101.96<,
840, - 101.96<, 860, - 101.96<, 880, - 101.96<, 898.1, - 101.96<<;
posts = ListPlot@points, PlotStyle ® PointSize@0.02DD;
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Group B
Show@pasture2, postsD
numberofposts = 23
23
postcost = 10 numberofposts
230
totalcost = postcost + fencecost
2077.68 dollars
Alex Bilzerian’s Method
Approach #3
This approach assumes that the cows will not leave the pasture, that the bull is the only animal
that needs to be enclosed, and that posts are not needed at every corner.
I began the problem by using a right triangle to contain an area that was approximately 1000 square meters
for the bull, using the river as the hypotenuse of the triangle. To save money on fencing, I only enclosed the
bull because it needed a smaller area and I placed the longest side of the triangular pen on the river.
COW AREA
150 M
page 5 of 13
PASTURE
that needs to be enclosed, and that posts are not needed at every corner.
I began the problem by using a right triangle to contain an area that was approximately 1000 square meters
B
for the bull, using the river as the hypotenuse of the Group
triangle.
To save money on fencing, I only enclosed
the
bull because it needed a smaller area and I placed the longest side of the triangular pen on the river.
12/14/13
COW AREA
PASTURE
150 M
RIVER
BULL-PEN...
70 M
29 M
75.78 M
120 M
After trial and error, I determined that a triangle with the side lengths 29, 70, and 75.78 would
enclose an area of 1015 square meters without having a very large perimeter or an impractical shape. The
extra 15 square meters were
added to the pen to ensure that there would be extra fencing for the construction process if needed.
areaofbullpen1 = H29 * 70L 2
1015
fencingneeded1 = 70 + 29
99
totalcost1 = Hfencingneeded1 * 5 L + HFloor@fencingneeded1 20.0D * 10L
535
totalcost1
535
The total cost of fencing for approach 1 was $535. This was determined by first finding the area of the
bull’s pen and making sure it was greater than 1000 square meters. I then proceeded to add together
the sides that were to be fenced, and multiplyed that length by $5.. To find the amount of neccesary
posts, I divided the fencing needed by 20 (1 post every 20 meters), rounded that value down to the
lowest integer using the Floor functon, and multiplyed that product by $10. Finnally, I added together
those two costs to get 535.
Approach #4
This approach evolved from approach 3 and assumes that the cows will not leave the pasture, that the
bull is the only animal that needs to be enclosed, and that posts are not needed at every corner.
This method uses half of a 50-sided regular polygon (pentacontagon) to enclose the bull to avoid the
cost associated with buying a post every 10 meters
on a curved surface. This method is not very practipage 6 of 13
cal, but it is extremely cost-effective when compared to approach 3. Below is an image of a pentacontagon. It mimics the shape of a circle which is important because circles are the best possible shapes
12/14/13
Group B
This approach evolved from approach 3 and assumes that the cows will not leave the pasture, that the
bull is the only animal that needs to be enclosed, and that posts are not needed at every corner.
This method uses half of a 50-sided regular polygon (pentacontagon) to enclose the bull to avoid the
cost associated with buying a post every 10 meters on a curved surface. This method is not very practical, but it is extremely cost-effective when compared to approach 3. Below is an image of a pentacontagon. It mimics the shape of a circle which is important because circles are the best possible shapes
for maximizing area and minimizing perimeter.
COW AREA
150 M
PASTURE
BULL_PEN...
RIVER
120 M
To make an even more cost-effective polygon, more sides could potentially used, but I
determined that 50 sides is sufficient since I round on meters of fencing needed to provide extra for the
construction process. The area of a pentacontagon is equal to approximately 198.682 times the sidelength squared.
For my the entire pentacontagon, I set the area equal to 2000 square meters so that half of it
would be equal to 1000 square meters for the bull.
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Group B

pentacontagon area
Result
25
s2 cotK
2
Π
50
O » 198.682 s2
Hassuming edge length sL
192.682 * sidelengthapproach2^ 2 = 2000
sidelengthapproach2^ 2 = 10.00663
sidelengthapproach2 = 3.17274 m
3.17274 * 25
3.17274 m
79.3185
80 * 5
400
400 + HFloor@80 20DL * 10
440
The total cost of fencing for approach 4 was $440. This was determined by first finding length of one of
the sides of the pentacontagon and then multiplying it by 25 (number of sides in half of the pentacontagon) to get about 80 meters of fencing. Then, I multiplyed that length by $5 to find the cost of the
fences themselves. To find the amount of neccesary posts, I divided the fencing needed by 20 (1 post
every 20 meters), rounded that value down to the lowest integer using the Floor functon, and multiplyed
that product by $10. Finnally, I added together those two costs to get 440.
As an extension of approach 2, I decided to try and use a quarter of a pentacontagon in an effort to see
if that would reduce the cost of enclosing the bull.
COW AREA
150 M
PASTURE
BULL_PEN...
RIVER
page 8 of 13
120 M
every 20 meters), rounded that value down to the lowest integer using the Floor functon, and multiplyed
that product by $10. Finnally, I added together those two costs to get 440.
12/14/13
Group B
As an extension of approach 2, I decided to try and use a quarter of a pentacontagon in an effort to see
if that would reduce the cost of enclosing the bull.
COW AREA
PASTURE
150 M
BULL_PEN...
RIVER
120 M
192.682 * sidelengthapproach3^ 2 = 4000
sidelengthapproach3^ 2 = 20.7596
sidelengthapproach3 = 4.55627 m
sidelengthapproach3 * 12.5 = 56.9534 m
H57 * 5L + HFloor@57 20D * 10L
305
$305 + cost of radius = total cost

radius of pentacontagon
Result
Π
1
s cscK
2
50
O » 7.96299 s
Hassuming edge length sL
radius = 18.1405 m
H18.1405 * 5L + HFloor@18.1405 20D * 10L
90.7025
305 + 90.9024
395.902
The total cost of fencing for the extension of approach 4 using a quarter of a pentacontagon was approximately $396. I found the cost of the 12.5 sides that made up the quarter of the pentacontagon and
added it to the cost of the radius of the pentacontagon to get $395.90.
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Group B
The total cost of fencing for the extension of approach 4 using a quarter of a pentacontagon was approximately $396. I found the cost of the 12.5 sides that made up the quarter of the pentacontagon and
added it to the cost of the radius of the pentacontagon to get $395.90.
Approach #5
Out of all of the approaches, method 5 is the most practical and realistic for Rancher Barney. This
approach assumes that both the bull and the cows need to be enclosed, and that posts are not needed
at every corner.
This approach uses a large trapazoid with a triangle within it to enclose both the cows and the bull
seperately. This is the most expensive method, yet it is under a different set of assumptions that cannot
be compared to the last 2 approaches.
COW AREA
PASTURE
150 M
BULL_PEN...
RIVER
120 M
Large Trapazoid Area = 11,000 square meters
Small Square Area = 1,000 square meters
Atrap = 1/2(120+b2) * 150 = 11,000
b2 = 26.66 meters... round up to 27 m
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Group B
x
x
h
b
b
Length of base for entire trapazoid = 120 m
Length of second base on top = 27 m
Height = 150 m
b = (120-27)/2
b = 23.25 m
x = Sqrt[h^2+b^2]
x = 151.791... round to 152 m
2x + b2 = 331
331 * 5 = $1815
Floor[331/20] *10 = $160
cost for trapazoid = $1975
b
h
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331 * 5 = $1815
Floor[331/20] *10 = $160
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Group B
cost for trapazoid = $1975
b
h
hypotenuse on river
Make height = 30
(30 * h)/2 - 1000
h = 66.66
Put height on trapazoid, put hypotenuse on river, pay for 30 m of fencing
(30 * $5) + Floor[30/20] *$10 = $160
160 + 1975 = $2135
Total cost for approach 5 = $2135
However this was not as low as the cost in approach 2.
Solution:
The best method under the assumption that both cows and bull have to be
fully enclosed is method 2 with a total cost of $2077.68. However if these assumptions are relaxed to
include solutions in which the cows and bull only have to be separated from each other and not
enclosed, then the most cost effective method is part 2 of approach 4 with a total cost of $395.90. But it
is a rather impractical method because of the short sides and many angles involved. With the same
assumptions, a more practical yet still relatively cost effective method would be approach 3 with a cost
of $535.
Generalizations:
There could be a change in fencing prices or post spac-
ing. Also, instead of being given an area for cows, there could be a number of cows and the area
needed could be found.
page 12 of 13
Self Assessment:
Our group did a good job of communicating various
12/14/13
Generalizations:
Group B
There could be a change in fencing prices or post spac-
ing. Also, instead of being given an area for cows, there could be a number of cows and the area
needed could be found.
Self Assessment:
Our group did a good job of communicating various
ideas and methods to each other to try and solve this problem. We came up with our own separate
ideas then presented them and fixed them, as well as extend them. Because we each did our own
method, responsibility was shared rather equally. When it came to doing the actual Mathematica presentation, because we ended up not using Juliana’s method because it was similar to Ashwin’s, she did the
writing portion. Over all, our group worked very well together.
page 13 of 13