• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

the fencing problem

Extracts from this document...

Introduction

The Fencing Problem

A farmer wants to fence a plot of level land and has exactly 1000 meters of fencing. The Farmer wants to have a perimeter of 1000m but is not concerned about the shape of the plot. I am going to investigate the maximum area of a land using different families of shapes. I will investigate all families of shapes and see which one has the maximum area. I will be using rectangles, triangles and polygons.

First I will be investigating Quadrilaterals and I will be using rectangles to find the maximum area using different lengths and keeping the perimeter the same.

image00.png

Perimeter= Width+Length+Width+Length

               = 250+250+250+250

               = 1000m

Area of Rectangle= Width*Length

                             = 250*250

                             = 62500m²

Width (m)

Lengths (m)

Perimeter (m)

Area (m²)

450

50

1000

22500

400

100

1000

40000

350

150

1000

52500

300

200

1000

60000

250

250

1000

62500

200

300

1000

60000

150

350

1000

52000

100

400

1000

40000

50

450

1000

22500

In my investigation I have found that the maximum area is 62500m² with the lengths 250 and 250. In my search I went down by 50 for the Width and up by 50 for the length so that the width and length add up to 1000 meters. think that this is not exactly the maximum area so I will refine my search by doing a decimal search.

Width (m)

Lengths (m)

Perimeter (m)

Area (m²)

249.5

250.5

1000

62499.75

249.6

250.4

1000

62499.84

249.7

250.3

1000

62499.91

249.8

250.2

1000

62499.96

249.9

250.1

1000

62499.99

250

250

1000

625000

After refining my search into a decimal search this table can tell me that the maximum area stayed the same at 62500m². I went up by 0.

...read more.

Middle

302.5

395

229.1

45252

1000

300

300

400

223.6

44721

1000

In this Search I went down by 2.5 for the lengths and up by 5 for the base

so that both lengths and base add up to 1000 meters. From this table I

can tell that in my first decimal search the maximum is 48110m², which I don’t think is as high as I could get it, therefore I will refine my search.

Length2(m)

Length2(m)

Base(m)

Height(m)

Area(m²)

Perimeter(m)

335

335

330

291.5

48105

1000

334.5

334.5

331

290.6

48109

1000

334

334

332

289.8

48111

1000

333.5

333.5

333

288.9

48112

1000

333

333

334

288.0

48112

1000

332.5

332.5

335

287.2

48110

1000

332

332

336

286.3

48107

1000

331.5

331.5

337

285.4

48103

1000

331

331

338

284.6

48098

1000

330.5

330.5

339

283.7

48091

1000

330

330

340

 282.8

48083

1000

In this Search I went down by 0.5 for the lengths and up by 1 for the base so that both the lengths and base add up to 1000 meters. From this table I can tell that there are two triangles with the same maximum area of 48112m². I will refine my search even further to find the exact maximum area for the triangles.

Length2(m)

Length2(m)

Base(m)

Height(m)

Area(m²)

Perimeter(m)

333

333

334

288.0

48096

1000

333.1

333.1

333.8

288.2

48100

1000

333.2

333.2

333.6

288.4

48105

1000

333.3

333.3

333.4

288.6

48109

1000

333.4

333.4

333.2

288.7

48097

1000

333.5

333.5

333

288.9

48101

1000

In this Search I went up by  0.1 for the lengths and down 0.2 for the base so that both the lengths and base add up to 1000 meters.

From this table I can say that the maximum area is 48109m² , with the lengths 333.3,333.3 and base 333.4. I will refine my search even further with two numbers behind the decimal.

Length2(m)

Length2(m)

Base(m)

Height(m)

Area(m²)

Perimeter(m)

333.31

333.31

333.38

288.63

48111.7347

1000

333.32

333.32

333.36

288.65

48112.182

1000

333.33

333.33

333.34

288.67

48112.6289

1000

333.34

333.34

333.32

288.66

48108.0756

1000

In this Search I went up by 0.01 for the lengths  and down 0.02 for the base so that both the lengths and base add up to 1000 meters.

After all my investigation of triangles I have found that the highest maximum area for the triangles is 48112m², with the lengths 333.33, 333.33 and base 333.34. The shape of this triangle is isosceles.

Next i will be investigating polygons, I will be investigating the pentagon, hexagon, heptagon, and octagon. I want to investigate these shapes because I think that after going through the rectangles and triangles and investigating them and their sides I have decided to do theses shapes as all of them have different number of sides.

Pentagon:

image09.png

image10.png

            200m

360°/5 = 72°

1 Triangle of Pentagon = 72°/ 2= 36°

Base = 1000/5

         = 200m

Half the base = 100m

Height of triangle=

                     Tan36°= 100/height  :cross multiply

                                = Height =100/Tan 36°

                                = 137.64m

Area of triangle = (Base* Height) /2

                          = (200*137.64)/2

                          = 13764m²

 Area of Pentagon =  Area of 1 triangle*5

                              = 13764*5

                              = 68820m²

In my investigation I have found that in the family of polygons the pentagon can give the maximum area of 68820m².

To investigate the n sides I have to first find a suitable formula which can help me find the area of a lot of n sides.

image11.png

image12.png

Tan (360/(2*n) = Tan(180/n)

Base = 1000/n

½Base = 1000/(n*2)

Tan(180/n) = (1000/2n)/h

H = (1000/2n)/Tan(180/n)

A = (1000/n)*((1000/2n)/Tan(180/n))/2

= (250000/n)*1/Tan (180/n)

Number of sides

Area

5

68819.09602

6

72168.78365

7

74161.47845

8

75444.17382

9

76318.81721

...read more.

Conclusion

Circle:

image06.png

Area of Circle = Πr²

Π = 3.14

Radius = (1000/Π)/2

            = 159.24m

Area of Circle = Πr²

                        = Π*(159.24)²

                        = 79622.16566m²

In my investigation I have found that the maximum area for the circles is 79622.16566m². I also found that the highest maximum area was for the circle after all the different shapes I investigated.

I have also noticed that the limit tot he area of the polygons is just a bit smaller than the maximum area of the circle.

Shape

Sides

Area (m²)

Triangle

3

62500

Quadrilaterals

4

48112

Pentagon

5

68820

Hexagon

6

72192

Heptagon

7

74249

Octagon

8

75448

Circle

79622

This table tells me that the highest maximum area is for the circle. After doing my investigation I went over the maximum areas for each family of shapes and compared them. For the Rectangles the maximum area I got was 62500m² which was a square. For the triangles the maximum area I got  was 48112m²I, which was a isosceles triangle.

For the polygons I used four different shapes which were a pentagon,  a hexagon, a heptagon and a octagon For the Pentagon the maximum area I got was 68820m². For the Hexagon the maximum area I got was 72192m². For the Heptagon the maximum area I got was 74249m². And for the octagon the maximum area I got was 75448m². I also did circles and found that the maximum area was 79622.16566m², which was the highest maximum area overall. I think the farmer should use the shape circle to fence his 1000 meter land.

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. The Fencing Problem

    Its base equals the base of the trapezium minus its top divided by two Base = (Base - Top) ( 2 Base = (250 - 150) ( 2 = 50m The hypotenuse of the triangle equals 100m Hypotenuse = 100m Base = 50m To find the height the equation is:

  2. Geography Investigation: Residential Areas

    (See Figure 8) To collect evidence of my investigation and what the residential area looks like I will take pictures to show what the different areas look like and if there is a problem with the area like bad traffic I have evidence to prove this.

  1. Maths Fence Length Investigation

    To find the length of the base of a segment I would divide 1000 by the number of sides, so I could put , but as I need to find half of that value I need to put . All The method that I used above has been put into an equation below.

  2. The Fencing Problem

    Angle (�) Height [h] (m) Perimeter (m) Area (m�) 100 400 10 69.46 1000 6945.93 100 400 20 136.81 1000 13680.81 100 400 30 200.00 1000 20000.00 100 400 40 257.12 1000 25711.50 100 400 50 306.42 1000 30641.78 100 400 60 346.41 1000 34641.02 100 400 70 375.88 1000 37587.70 100 400 80 393.92 1000 39392.31 100 400

  1. Fencing Problem

    * I will multiply the base x height to give me the area of the two triangles. The reason I am not using 1/2 base x height is because that will find me the answer of one of the triangles, i.e.

  2. My investigation is about a farmer who has exactly 1000 metres of fencing and ...

    I will plot this information as well as the information from the other graph onto a new graph showing number of sides against area: This graph shows that my prediction is still correct as the area is getting larger as the amount of sides increase.

  1. Fencing problem.

    � 331 = 55939 m2 Trapezium The next shape that I shall be investigating is the trapezium. I have shown the formula to find the area of a trapezium below: Area of trapezium = 1/2 � (a + b) � h The height of the trapezium is unknown and it would have to be obtained before proceeding.

  2. Investigate the shapes that could be used to fence in the maximum area using ...

    To find the length of the base of a segment I would divide 1000 by the number of sides, so I could put , but as I need to find half of that value I need to put . All The method that I used above has been put into an equation below.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work