# The Fencing Problem

Extracts from this document...

Introduction

Mike Kounoupias

I am investigating a problem to do with a plot of land with either a perimeter or a circumference of 1000 m. There are many possibilities for the she farmer to plot her fence round the area of land. She can plot in triangles, hexagons, decagons, rectangles, squares etc.

First, I am going to try to find out the greatest area for a rectangle. If I try using a rectangle with length of 400 m, and a width of 100 m, I am going to find out its area. Here, I have a rectangle of width w, which must have a length of 500 – w, if the perimeter is to 1000 m.

CHECK:

W + 500 – w + 500 – w + 500 – w.

= 500 w – w2

A = 500 w – w 2

Its area is w (500 – w)

A = 500w - w2

Middle

Area

22500

40000

52500

60000

62500

60000

52500

40000

I shall draw a graph and find the value of w for which A is greatest.

The graph below shows me that a rectangle with a perimeter of 1000 m has a maximum area when it becomes a square, having sides of 250 m. The area is then 62 500 m2. I am now going to do the same with a triangle. I will find out the greatest area for a triangle.

Area of Triangle = ½ Base * Height

a= ½ 200 * 300 = 30 000 m

This is much less than the area of a square. Perhaps if I make the base and the height equal, the result will improve.

Pythagoras gives y = x + x.

y = 2x = 2x

P = x + x + 2x

2x + 2x = 1000

x ( 2 + 2 ) = 1000

x = 1000 / 2 + 2~ 292.9 M

A = 1/2 * 292.9 * 292.9

= 42, 893.2 m

Would an equilateral triangle be better than a right angled triangle, since it has 3 equal sides?

Conclusion

Reg Polygon of “n” sides

1000/h = length of side

180 / ntan (180/n) = 500/n/h

h = 500 /n / tan(180/n)

Area of reg. polygon with n sides

= n * 500 / n * 500 / n / tan (180 / n ) = 250,000/ n TAN (180 / n )

I shall use a table to test my formula for various values of n.

n | Area |

3 | 250 000/ 3 tan 60 = 48112.523 |

5 | 250 000/ 5 tan 36 = 68819.096 |

10 | 350 000/10 tan18 = 76942. 088 |

These three results confirm that my general formula is true, because the values agree with the previous areas.

100 | 250 000/ 100tan 1.8 = 79551.289 |

1000 | 250 000/ 1000 tan .18 = 79577.209 |

10000 | 350 000/10000 tan.018 = 79577.468 |

A circle is very close to a regular polygon having a very large number of sides. What is the area of a circle with perimeter of 1000 m ? I will calculate this value.

P = 1000 = 2πr

R = 1000/2π = 500/π

= 159.1549431 m

Area = πr2.

Π(159.1549431)2

= 79577.47155 m2

Conclusion

For a given length of fencing (such as 1000 m ) the greatest possibleb area which the farmer could enclose would be given by a circle pf radius 159.1549431 m,; the biggest shape it could be is a circle.

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

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