Maths Coursework - The Fencing Problem
Maths Coursework
The Fencing Problem
By Anil Vekaria
0H
Aim:
A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate different shapes the fencing can make to achieve the largest area.
I am going to start investigating different shape rectangles because they are the easiest shapes to work put the perimeter all of these shapes will have a perimeter of 1000 metres. Below are 2 rectangles (to scale) showing how different shapes with the same perimeter can have different areas. I will use a scale of 1cm:100m.
) 2)
400 metres (4 centimetres)
Height 300m(3cm)
200m (2cm)
00 metres (1 centimetre)
Width
I will work out the area of both rectangles by using the formula below. Both rectangles have a perimeter of 1000m
)
Area of rectangle = height * width
Area of rectangle = 400m * 100m
Area of rectangle = 40000 m²
2)
Area of rectangle = height * width
Area of rectangle = 300m * 200m
Area of rectangle = 60000m²
As you can notice the areas of the rectangles 1 and 2 are different though the perimeters of both are 1000m.
Now I will put the areas widths and lengths of rectangles. I will change the value of the widths and go up in increments of 10m.
I will not use negative numbers for they are realistically impossible. Mathematically negative lengths are possible but I know that investigating this wont give me the answers I want.
I will put my results in a table now.
Width
Length
Difference between W and L
Perimeter =SUM(A:A+B:B)*2
Area=SUM(A:A*B:B)
0
500
500
000
0
0
490
480
000
4900
20
480
460
000
9600
30
470
440
000
4100
40
460 studentcentral.co.uk
420
000
8400
50
450
400
000
22500
60
440
380
000
26400
70
430
360
000
30100
80
420
340
000
33600
90
410
320
000
36900
00
400
300
000
40000
10 wwff ffw stffffud eff ffnt cff enfftral ffcoff uk!
390
280
000
42900
20
380
260
000
45600
30
370
240
000 wwbb bbw stbbbbud ebb bbnt cbb enbbtral bbcobb uk.
48100
40
360
220
000
50400
50
350
200
000
52500
60
340
80
000
54400
70
330 wwfe few stfefeud efe fent cfe enfetral fecofe uk!
60
000
56100
80
320
40
000
57600
90
310
20
000
58900
200
300
00
000
60000
210
290
80
000
60900
220
280
60
000
61600
230
270
40
000 wwcf cfw stcfcfud ecf cfnt ccf encftral cfcocf uk.
62100
240
260
20
000
62400
250
250
0
000
62500
260
240
20
000
62400
270
230
40
000
62100
280
220
60
000
61600
290
210
80
000
60900 lj3 Visit studentcentral ag co ag uk ag for more ag Do not ag redistribute lj3
300
200
00
000
60000
310
90
20
000
58900
320
80
40
000
57600
330
70
60
000
56100
340
60
80
000
54400
350
50
200
000
52500
360
40
220 wwee eew steeeeud eee eent cee eneetral eecoee uk!
000
50400
370
30
240
000
48100
380
20
260
000
45600
390
10
280
000
42900
400
00
300
000
40000
410
90
320
000
36900
420
80
340
000
33600 wwcf cfw stcfcfud ecf cfnt ccf encftral cfcocf uk.
430
70
360
000
30100
440
60
380
000
26400
450
50
400
000
22500
460
40
420
000
8400
470
30
440
000
4100
480
20
460
000
9600
490
0
480
000
4900
500
0
500
000
0
The highlighted row gives the biggest area; after I go past this row I start to repeat my self.
A square of perimeter 1000m gives me the largest area. I will further investigate this because I was going up in increments of 10. I will go into the decimal widths and lengths.
Below is a table of results.
Width
Length
Area
249
251
62499
249.5
250.5
62499.8
24975
250.25
6249994
250
250
62500
This coursework from www.studentcentral.co.uk (http://www.studentcentral.co.uk/coursework/essays/2257.html)
Reproduction or retransmission in whole or in part expressly prohibited
250.25
249.75
62499.9
250.5
249.5
62499.8
251
249
62499
From this table I can see that the perfect square of perimeter 1000m produces the largest area.
From this graph I can see that the highest point is on the 250m mark. The graph is a parabola and is symmetrical. At the centre point of the graph I reach the highest value for the area, after this I just repeat my self.
In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m * 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side. To work out the area of a rectangle with a base length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.
Area = x(500 - x)
Key: x = width of rectangle
All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas. wwbc bcw stbcbcud ebc bcnt cbc enbctral bccobc uk;
Now I will investigate triangles. I will first investigate isosceles triangles because triangles such as scalene have more than one different variable so there are millions of possible combinations. If I know the base length I can work out the length of ...
This is a preview of the whole essay
Area = x(500 - x)
Key: x = width of rectangle
All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas. wwbc bcw stbcbcud ebc bcnt cbc enbctral bccobc uk;
Now I will investigate triangles. I will first investigate isosceles triangles because triangles such as scalene have more than one different variable so there are millions of possible combinations. If I know the base length I can work out the length of the other two sides because they are the same.
For example if the base is 200m. I can take this away from 1000m and that answer I can divide by 2. Or more simply:
Side = (1000 - 200) ÷ 2 = 400
To work out the area I need to know the height of the triangle. To work out the height I can use Pythagoras' theorem. Below is the formula and area when using a base of 200m.
H2 = h2 - a2
H2 = 4002 - 1002
H2 = 150000
H = 387.298
/2 * 200 * 387.298 = 38729.833m.
On the next page there is a table of results for triangles.
Base (m)
Side (m)
Height (m)
Area (m)
0
500.0
500.000
0.000
0
495.0
494.975
2474.874
20
490.0
489.898
4898.979
30 wwad adw stadadud ead adnt cad enadtral adcoad uk.
485.0
484.768
7271.520
40
480.0
479.583
9591.663
50
475.0
474.342
1858.541
60
470.0
469.042
4071.247
70
465.0
463.681
6228.832
80
460.0
458.258
8330.303
90
455.0
452.769
20374.617
00
450.0
447.214
22360.680
10
445.0
441.588
24287.342
20
440.0
435.890
26153.394
30
435.0
430.116
27957.557
40
430.0
424.264
29698.485
50
425.0
418.330
31374.751
60
420.0
412.311
32984.845
70
415.0
406.202
34527.163
80
410.0
400.000
36000.000
90
405.0
393.700
37401.537
200
400.0
387.298
38729.833
210
395.0 from www.studentcentral.co.uk
380.789
39982.809
220
390.0
374.166
41158.231
230
385.0
367.423
42253.698
240
380.0
360.555
43266.615
250
375.0
353.553
44194.174
260
370.0
346.410
45033.321
270
365.0
339.116
45780.727
280
360.0
331.662
46432.747
290
355.0
324.037
46985.370
300 wwba baw stbabaud eba bant cba enbatral bacoba uk!
350.0
316.228
47434.165
310
345.0
308.221
47774.209
320
340.0
300.000
48000.000
330
335.0
291.548
48105.353
333.3
333.3
288.675
48112.522
340
330.0
282.843
48083.261 wwdg dgw stdgdgud edg dgnt cdg endgtral dgcodg uk.
350
325.0
273.861
47925.724
360
320.0
264.575
47623.524
370
315.0
254.951
47165.931
380
310.0 wwda daw stdadaud eda dant cda endatral dacoda uk!
244.949
46540.305
390
305.0
234.521
45731.554
400
300.0
223.607
44721.360
410
295.0
212.132
43487.067
420
290.0
200.000
42000.000
430
285.0
87.083
40222.817
440
280.0
73.205
38105.118
450
275.0
58.114
35575.624
460
270.0
41.421
32526.912
470
265.0
22.474
28781.504
480
260.0
00.000
24000.000
490
255.0
70.711
7324.116
500
250.0
0.000
0.000
From the table I can see that the regular shape of the family has given me the largest area.
I can see that the regular shapes of these families give the largest area.
Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on. This would also be a lot easier as many of the other shapes have millions of different variables.
The next shape that I am going to investigate is the pentagon.
Because there are 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle, with the top angle being 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540 each. Because every isosceles triangle can be split into 2 equal right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle is 100m.
Tanq = opposite/adjacent
Tan54 = o / a
O = tan54 * a
O = tan54 *100
O = 137.638192m
37.638192m * 200 =27527.63841 s97VJf4i from s97VJf4i student s97VJf4i central s97VJf4i co s97VJf4i uk
27527.63841/2=13763.81921
3763.81921*5=68819.096m²
All of the results that I have got so far have shown that as the number of side's increases, the area increases. I am going to investigate this further with a regular hexagon (6 sides) and a regular heptagon (7 sides).
I am going to work out the area of the 2 shapes using the same method as before.
Hexagon:
000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.
360 ÷ 6 = 60 ÷ 2 = 30
Area = 1/2 * b * H = 1/2 * 83 1/3 * 144.338 = 6014.065
6014.065 * 12 = 72168.784m2
Heptagon:
000 ÷ 7 = 142.857 ÷ 2 = 71.429m
360 ÷ 7 = 51.429 ÷ 2 = 25.714 degrees
Area = 1/2 * b * H = 1/2 * 71.429 * 148.323 = 5297.260
5297.260 * 14 = 74161.644m2
My predictions were correct and as the number of side's increases, the area increases. Below is a table of the number of sides against area
No. Of sides
Area (m2)
3
48112.522
4
62500.000
5
68819.096
6
72168.784
7
74161.644 wwga gaw stgagaud ega gant cga engatral gacoga uk;
To find the general formula for an n sided regular polygon I am going to construct a formula.
To find the length of the base of a segment I would divide 1000 by the number of sides, so I could use 1000/s (s=number of sides), but because I want to find half of one segment I use (1000/s)/2. To find the interior angle of the n sided polygon I can use, (360/s)/2. I am finding half of the interior angle of a segment. Then I have to find the length of the perpendicular in the segment. To do this I use the tangent formula. Adjacent = opposite/ Tanq
A = [(1000/n)/2]/ Tan (360/s)/2
Then to find the area of the regular polygon I have to multiply the value of the line A by half of the base (1000/s)/2 then I have to multiply that by the number of sided the regular polygon has. So basically it's A * base *0.5 * number of sides.
As the table above shows as the number of sides go up the area goes up. Now I am going to investigate the area of a perfect circle. I predict that this will ultimately have the largest area because it has an infinite amount of sides.
Diameter 318.47m
pR²
3.14 * 159.235 (2 d.p) * 159.235 (2 d.p) = 79617.16561m²
I can see that this produces the largest area.
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Maths Coursework - The Fencing Problem
Maths Coursework
The Fencing Problem
By Anil Vekaria
0H
Aim:
A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the biggest size. I will investigate different shapes the fencing can make to achieve the largest area.
I am going to start investigating different shape rectangles because they are the easiest shapes to work put the perimeter all of these shapes will have a perimeter of 1000 metres. Below are 2 rectangles (to scale) showing how different shapes with the same perimeter can have different areas. I will use a scale of 1cm:100m.
) 2)
400 metres (4 centimetres)
Height 300m(3cm)
200m (2cm)
00 metres (1 centimetre)
Width
I will work out the area of both rectangles by using the formula below. Both rectangles have a perimeter of 1000m
)
Area of rectangle = height * width
Area of rectangle = 400m * 100m
Area of rectangle = 40000 m²
2)
Area of rectangle = height * width
Area of rectangle = 300m * 200m
Area of rectangle = 60000m²
As you can notice the areas of the rectangles 1 and 2 are different though the perimeters of both are 1000m.
Now I will put the areas widths and lengths of rectangles. I will change the value of the widths and go up in increments of 10m.
I will not use negative numbers for they are realistically impossible. Mathematically negative lengths are possible but I know that investigating this wont give me the answers I want.
I will put my results in a table now.
Width
Length
Difference between W and L
Perimeter =SUM(A:A+B:B)*2
Area=SUM(A:A*B:B)
0
500
500
000
0
0
490
480
000
4900
20
480
460
000
9600
30
470
440
000
4100
40
460 studentcentral.co.uk
420
000
8400
50
450
400
000
22500
60
440
380
000
26400
70
430
360
000
30100
80
420
340
000
33600
90
410
320
000
36900
00
400
300
000
40000
10 wwff ffw stffffud eff ffnt cff enfftral ffcoff uk!
390
280
000
42900
20
380
260
000
45600
30
370
240
000 wwbb bbw stbbbbud ebb bbnt cbb enbbtral bbcobb uk.
48100
40
360
220
000
50400
50
350
200
000
52500
60
340
80
000
54400
70
330 wwfe few stfefeud efe fent cfe enfetral fecofe uk!
60
000
56100
80
320
40
000
57600
90
310
20
000
58900
200
300
00
000
60000
210
290
80
000
60900
220
280
60
000
61600
230
270
40
000 wwcf cfw stcfcfud ecf cfnt ccf encftral cfcocf uk.
62100
240
260
20
000
62400
250
250
0
000
62500
260
240
20
000
62400
270
230
40
000
62100
280
220
60
000
61600
290
210
80
000
60900 lj3 Visit studentcentral ag co ag uk ag for more ag Do not ag redistribute lj3
300
200
00
000
60000
310
90
20
000
58900
320
80
40
000
57600
330
70
60
000
56100
340
60
80
000
54400
350
50
200
000
52500
360
40
220 wwee eew steeeeud eee eent cee eneetral eecoee uk!
000
50400
370
30
240
000
48100
380
20
260
000
45600
390
10
280
000
42900
400
00
300
000
40000
410
90
320
000
36900
420
80
340
000
33600 wwcf cfw stcfcfud ecf cfnt ccf encftral cfcocf uk.
430
70
360
000
30100
440
60
380
000
26400
450
50
400
000
22500
460
40
420
000
8400
470
30
440
000
4100
480
20
460
000
9600
490
0
480
000
4900
500
0
500
000
0
The highlighted row gives the biggest area; after I go past this row I start to repeat my self.
A square of perimeter 1000m gives me the largest area. I will further investigate this because I was going up in increments of 10. I will go into the decimal widths and lengths.
Below is a table of results.
Width
Length
Area
249
251
62499
249.5
250.5
62499.8
24975
250.25
6249994
250
250
62500
This coursework from www.studentcentral.co.uk (http://www.studentcentral.co.uk/coursework/essays/2257.html)
Reproduction or retransmission in whole or in part expressly prohibited
250.25
249.75
62499.9
250.5
249.5
62499.8
251
249
62499
From this table I can see that the perfect square of perimeter 1000m produces the largest area.
From this graph I can see that the highest point is on the 250m mark. The graph is a parabola and is symmetrical. At the centre point of the graph I reach the highest value for the area, after this I just repeat my self.
In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m * 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side. To work out the area of a rectangle with a base length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.
Area = x(500 - x)
Key: x = width of rectangle
All of these results fit into the graph line that I have, making my graph reliable. Also the equation that I used is a quadratic equation, and all quadratic equations form parabolas. wwbc bcw stbcbcud ebc bcnt cbc enbctral bccobc uk;
Now I will investigate triangles. I will first investigate isosceles triangles because triangles such as scalene have more than one different variable so there are millions of possible combinations. If I know the base length I can work out the length of the other two sides because they are the same.
For example if the base is 200m. I can take this away from 1000m and that answer I can divide by 2. Or more simply:
Side = (1000 - 200) ÷ 2 = 400
To work out the area I need to know the height of the triangle. To work out the height I can use Pythagoras' theorem. Below is the formula and area when using a base of 200m.
H2 = h2 - a2
H2 = 4002 - 1002
H2 = 150000
H = 387.298
/2 * 200 * 387.298 = 38729.833m.
On the next page there is a table of results for triangles.
Base (m)
Side (m)
Height (m)
Area (m)
0
500.0
500.000
0.000
0
495.0
494.975
2474.874
20
490.0
489.898
4898.979
30 wwad adw stadadud ead adnt cad enadtral adcoad uk.
485.0
484.768
7271.520
40
480.0
479.583
9591.663
50
475.0
474.342
1858.541
60
470.0
469.042
4071.247
70
465.0
463.681
6228.832
80
460.0
458.258
8330.303
90
455.0
452.769
20374.617
00
450.0
447.214
22360.680
10
445.0
441.588
24287.342
20
440.0
435.890
26153.394
30
435.0
430.116
27957.557
40
430.0
424.264
29698.485
50
425.0
418.330
31374.751
60
420.0
412.311
32984.845
70
415.0
406.202
34527.163
80
410.0
400.000
36000.000
90
405.0
393.700
37401.537
200
400.0
387.298
38729.833
210
395.0 from www.studentcentral.co.uk
380.789
39982.809
220
390.0
374.166
41158.231
230
385.0
367.423
42253.698
240
380.0
360.555
43266.615
250
375.0
353.553
44194.174
260
370.0
346.410
45033.321
270
365.0
339.116
45780.727
280
360.0
331.662
46432.747
290
355.0
324.037
46985.370
300 wwba baw stbabaud eba bant cba enbatral bacoba uk!
350.0
316.228
47434.165
310
345.0
308.221
47774.209
320
340.0
300.000
48000.000
330
335.0
291.548
48105.353
333.3
333.3
288.675
48112.522
340
330.0
282.843
48083.261 wwdg dgw stdgdgud edg dgnt cdg endgtral dgcodg uk.
350
325.0
273.861
47925.724
360
320.0
264.575
47623.524
370
315.0
254.951
47165.931
380
310.0 wwda daw stdadaud eda dant cda endatral dacoda uk!
244.949
46540.305
390
305.0
234.521
45731.554
400
300.0
223.607
44721.360
410
295.0
212.132
43487.067
420
290.0
200.000
42000.000
430
285.0
87.083
40222.817
440
280.0
73.205
38105.118
450
275.0
58.114
35575.624
460
270.0
41.421
32526.912
470
265.0
22.474
28781.504
480
260.0
00.000
24000.000
490
255.0
70.711
7324.116
500
250.0
0.000
0.000
From the table I can see that the regular shape of the family has given me the largest area.
I can see that the regular shapes of these families give the largest area.
Because the last 2 shapes have had the largest areas when they are regular, I am going to use regular shapes from now on. This would also be a lot easier as many of the other shapes have millions of different variables.
The next shape that I am going to investigate is the pentagon.
Because there are 5 sides, I can divide it up into 5 segments. Each segment is an isosceles triangle, with the top angle being 720. This is because it is a fifth of 360. This means I can work out both the other angles by subtracting 72 from 180 and dividing the answer by 2. This gives 540 each. Because every isosceles triangle can be split into 2 equal right-angled triangles, I can work out the area of the triangle, using trigonometry. I also know that each side is 200m long, so the base of the triangle is 100m.
Tanq = opposite/adjacent
Tan54 = o / a
O = tan54 * a
O = tan54 *100
O = 137.638192m
37.638192m * 200 =27527.63841 s97VJf4i from s97VJf4i student s97VJf4i central s97VJf4i co s97VJf4i uk
27527.63841/2=13763.81921
3763.81921*5=68819.096m²
All of the results that I have got so far have shown that as the number of side's increases, the area increases. I am going to investigate this further with a regular hexagon (6 sides) and a regular heptagon (7 sides).
I am going to work out the area of the 2 shapes using the same method as before.
Hexagon:
000 ÷ 6 = 166 1/6 ÷ 2 = 83 1/3.
360 ÷ 6 = 60 ÷ 2 = 30
Area = 1/2 * b * H = 1/2 * 83 1/3 * 144.338 = 6014.065
6014.065 * 12 = 72168.784m2
Heptagon:
000 ÷ 7 = 142.857 ÷ 2 = 71.429m
360 ÷ 7 = 51.429 ÷ 2 = 25.714 degrees
Area = 1/2 * b * H = 1/2 * 71.429 * 148.323 = 5297.260
5297.260 * 14 = 74161.644m2
My predictions were correct and as the number of side's increases, the area increases. Below is a table of the number of sides against area
No. Of sides
Area (m2)
3
48112.522
4
62500.000
5
68819.096
6
72168.784
7
74161.644 wwga gaw stgagaud ega gant cga engatral gacoga uk;
To find the general formula for an n sided regular polygon I am going to construct a formula.
To find the length of the base of a segment I would divide 1000 by the number of sides, so I could use 1000/s (s=number of sides), but because I want to find half of one segment I use (1000/s)/2. To find the interior angle of the n sided polygon I can use, (360/s)/2. I am finding half of the interior angle of a segment. Then I have to find the length of the perpendicular in the segment. To do this I use the tangent formula. Adjacent = opposite/ Tanq
A = [(1000/n)/2]/ Tan (360/s)/2
Then to find the area of the regular polygon I have to multiply the value of the line A by half of the base (1000/s)/2 then I have to multiply that by the number of sided the regular polygon has. So basically it's A * base *0.5 * number of sides.
As the table above shows as the number of sides go up the area goes up. Now I am going to investigate the area of a perfect circle. I predict that this will ultimately have the largest area because it has an infinite amount of sides.
Diameter 318.47m
pR²
3.14 * 159.235 (2 d.p) * 159.235 (2 d.p) = 79617.16561m²
I can see that this produces the largest area.
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