# The Fencing Problem.

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Introduction

Edexcel – 1998 Specimen Coursework Task Syllabus 1385

Mathematics GCSETHE FENCING PROBLEMTiers F + I + H

A farmer has exactly 1000 metres of fencing; with it she wishes to fence off a plot of land.

She is not concerned about the shape of the plot, but it must have a perimeter of 1000m. So it could be

or anything else with a perimeter (or circumference) of 1000m

What she does wish to do is fence off the plot of land which contains the maximum area.

Investigate the shape, or shapes of the plot of land which have a maximum area.

Throughout this investigation I will check that the perimeter is 1000 meters by finding the total of all the outer sides. Also I will use refining as a way of finding the maximum area. When I talk about using the maximum area of the previous table the maximum area of each table will be highlighted.

## Rectangles

The first shape I will test will be a rectangle. Having been told that the perimeter must be 1000 meters I will find the areas of three rectangles, each with different lengths of sides, making sure that the perimeter is kept the same.

To calculate the area I will use the formula LENGTH x WIDTH = AREA

or Area = lw.

Rectangle A: l = 450m

w = 10m

Area = 450 x 10

Area = 4500m2

Rectangle B: l = 300m

w = 200m

Area = 300 x 200

Area = 60000m2

Rectangle C: l = 100m

w = 400m

Area = 100 x 400

Area = 40000m2

Having carried out the above calculations I will create a spreadsheet with formulae to carry out more calculations.

Middle

s = 300m

b/2 = 200m

h = 3002-1002

h = √50000m

h = 223.6068m

Area = 400 x 223.6068 / 2

Area = 44721.35955m2

Triangle C: b = 200m

s = 400m

b/2 = 100m

h = 4002-1002

h = √150000m

h = 387.29833m

Area = 200 x 387.29833 / 2

Area = 38729.38466m2

After completing the above tests I will create a spreadsheet with formulae to carry out more calculations. The headings will consist of Base, 1 equal side, Perimeter, Height and Area. Under the base heading there will be a variable number between 1 and 500. The first formula will be used to calculate the length of one equal side of the isosceles triangle. The formula will be =(1000-B2)/2 where B2 is the base. It will be divided by 2 because 1000-B2 would give the sum of the two equal sides together. As previously , for the rectangles, there will be a formula to check that the perimeter is 1000m. This will be the base plus, one equal side multiplied by two or =B2+(C2*2). The main formula in this spreadsheet will be the one used to find the height. In a spreadsheet there are codes that represent calculations carried out. These are put at the front of the formula and the substitute for square root is SQRT. So my formula will be the square root of 1 equal side squared, minus half the base squared. However before entering my formula I found out that using the power sign (^) doesn’t give accurate results and in order to square numbers I must multiply the number by itself instead of using such a sign. Therefore the formula entered into the spreadsheet will be

=SQRT((C2*C2)

Conclusion

Hypothesis

I predict that a circle will give the largest area because of my tests on regular polygons. I also predict that the maximum area given will be pretty close to that of a regular polygon with 30 sides (79286.37045m2) because of the curve on the graph plotted for the regular polygon section.

To find the area of a circle I will be required to use the formulae 2πr and πr2. The circumference must be 1000m and before finding the area I need to find the radius.

Radius = (1000/2)/π

r = 500/π

r = 159.1549431m

Area = π*159.15494312

Area = 79577.47155m2

To complete this in a spreadsheet under the circumference heading I would enter 1000. Under the radius heading I would use the formula =(C2/2)/PI() where C2 is the circumference. Finally under the Area heading I would enter the formula =PI()*(D2*D2) where D2 is the radius. The headings and formulas will be entered as shown in the table below.

Number of Sides | Circumference (m) | Radius (m) | Area (square m) | |

Infinite | 1000 | =(C2/2)/PI() | =PI()*(D2*D2) | |

Formula | - | 2πr | (Circumference/2)/π | πr2 |

## Proof

Number of Sides | Circumference (m) | Radius (m) | Area (square m) |

Infinite | 1000 | 159.1549431 | 79577.47155 |

The table above clearly proves my hypothesis correct. The working out also proves my hypothesis correct.

## Conclusion

Having completed the spreadsheet table I can conclude that a circle gives the maximum area and that the result was close to that given by a 30 sided regular polygon. A circle provides the maximum area possible for fencing of length 1000m. The maximum area possible is: -

79577.47155m2

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