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• Level: GCSE
• Subject: Maths
• Word count: 3635

# The Fencing Problem

Extracts from this document...

Introduction

The Fencing Problem

A farmer has exactly 1000 metres of fencing and wants to fence off a plot of level land.

She is not concerned about the shape of the plot, but it must have a perimeter (or circumference) of 1000m.

She wishes to fence off the plot of land, which contains the maximum area.

I am going to investigate the shape or, shapes, that have the maximum area with a perimeter of a 1000 metres.  I am going to first investigate triangles, because they are polygons with the least number of sides. Then I will progress to other polygons with increasing number of sides.  Finally, I will work out the area of a polygon with infinite number of sides, i.e. a circle, with the circumference of 1000m.

Triangles

I will investigate isosceles triangles because triangles such as scalene have more than one different variable so there are lots of possible combinations. If I know the length of the base of an isosceles triangle, I can work out the lengths of the other two sides because they are the equal.  For example, if the base is 200m, the sides would be 400m long.  To work this out I used the following formula:

length of side = (1000-base )/2

This is a diagram showing the lengths of a triangle.

b = base of triangle

Middle

333.4

333.30

288.617

48112.520

333.5

333.25

288.531

48112.504

333.6

333.20

288.444

48112.476

333.7

333.15

288.357

48112.435

333.8

333.10

288.271

48112.381

333.9

333.05

288.184

48112.314

334.0

333.00

288.097

48112.233

Using this results table I can draw a final graph of area against base of triangle. From the graph, I can see that the maximum area of a triangle with a range of bases, ranging from 333m to 334m, is a triangle with a base of 333.3m, with sides of 333.3m long and a perimeter of 1000m.The area of this triangle is 48112.522m².  The triangle is an equilateral triangle.

I will investigate if and how the interior angle of a parallelogram effects the area of the parallelogram and see if the parallelogram has a bigger area than a rectangle with the same perimeter of 1000m.  To work out the area of a parallelogram, I will use the following formula:

l = length of parallelogram

w = width of parallelogram

h = perpendicular height of parallelogram

a = area of parallelogram

a = l * h

However, because I only know the length and width of the parallelograms, I will need to work out the perpendicular height.  For example, if the length of a parallelogram was 300m and the width of a parallelogram was 200m, I wouldn't be able to use the formula shown above. The formula to work out the perpendicular height is:

h = w * sinx°

I used sine because I know the hypotenuse (width) and I want to work out the opposite (perpendicular height). I will now work out the area of the following parallelogram.

h         = 200m * sin45°

= 141.4m

a        = 300m * 141.4m

= 42420m²

I will change the angle x but keep the length and width the same to find the interior angle needed to create the maximum area.  The results were as followed:

 x° height/m area/m² 10 34.73 10418.89 20 68.40 20521.21 30 100.00 30000.00 40 128.56 38567.26 50 153.21 45962.67 60 173.21 51961.52 70 187.94 56381.56 80 196.96 59088.47 90 200.00 60000.00 100 196.96 59088.47 110 187.94 56381.56 120 173.21 51961.52 130 153.21 45962.67 140 128.56 38567.26 150 100.00 30000.00 160 68.40 20521.21 170 34.73 10418.89

These results show that a parallelogram has the greatest area when angle x is 90° i.e. when it is a rectangle.  Therefore, a rectangle has a greater area than a parallelogram.

I will investigate the areas of different rectangles, that each have the same perimeter of 1000 metres, because a rectangle is an easier shape for which to work out the area for than any other quadrilaterals.  To work out the area, I will use the following formula:

l = length of rectangle

w = width of rectangle

a = area of rectangle

a = l * w

The two examples I worked out are shown below.

The area of the rectangles above are as follows.

1. area = length * width

area = 150m * 350m

area = 525000m²

1. area = length * width

area = 200m * 300m

area = 60000m²

The examples above show that not all rectangles with the same perimeter have the same area.

The formula to work out the area of any rectangle, with a perimeter of 1000m for any given length is:

width        = 500 - length

area        = length * width

Substituting the width in the formulas above creates this formula:

area        = length * (500 - length)

area        = (500 * length) - length²

I'm going to use the formula shown above to find the maximum area of a rectangle with a perimeter of 1000 metres.  To save time and go through all the possible measurements by a calculator, I'm going to make a spreadsheet.  I will change the length of the rectangle in steps of 10m.  The formulas I will use in the spreadsheet are:

width        = 500 - length

area        = length * width

The results are as followed:

 length/m width/m area/m² 0 500 0 10 490 4900 20 480 9600 30 470 14100 40 460 18400 50 450 22500 60 440 26400 70 430 30100 80 420 33600 90 410 36900 100 400 40000 110 390 42900 120 380 45600 130 370 48100 140 360 50400 150 350 52500 160 340 54400 170 330 56100 180 320 57600 190 310 58900 200 300 60000 210 290 60900 220 280 61600 230 270 62100 240 260 62400 250 250 62500 260 240 62400 270 230 62100 280 220 61600 290 210 60900 300 200 60000 310 190 58900 320 180 57600 330 170 56100 340 160 54400 350 150 52500 360 140 50400 370 130 48100 380 120 45600 390 110 42900 400 100 40000 410 90 36900 420 80 33600 430 70 30100 440 60 26400 450 50 22500 460 40 18400 470 30 14100 480 20 9600 490 10 4900 500 0 0

Conclusion

a = n * ((500/n) * ((500/n) * tan(90-(180/n)))

Simplified:

a = 500 * ((500/n) * tan(90-(180/n)))

Multiplied out brackets:

a = 250000/n * tan(90-(180/n)))

I used the formulas above to see what happens as you increase the number of sides.  The results are as followed:

 Number of sides area/m² 10 76942.09 20 78921.89 30 79286.37 40 79413.78 50 79472.72 60 79504.74 70 79524.04 80 79536.56 90 79545.15 100 79551.29 110 79555.83 120 79559.29 130 79561.98 140 79564.11 150 79565.84 160 79567.24 170 79568.41 180 79569.39 190 79570.22 200 79570.93 210 79571.53 220 79572.06 230 79572.52 240 79572.93 250 79573.28 260 79573.60 270 79573.88 280 79574.13 290 79574.36 300 79574.56

From this results table, I can produce a graph of area against number of sides.  From the graph and table, I can see that as you increase the number of sides, the area increase.  Therefore, the shape that has the maximum area, with a perimeter of 1000m, is a circle because it has infinite sides.  As n increases and approaches infinite, the shape should become a circle.  Now I am going to investigate a circle and see if this is true.

Circle

a = area of circle

c = circumference of circle

a         =

c        = 2r

If the circumference was 1000m:

1000m        = 2r

1000/2        = r

Simplified:

500/        = r

Substitute r in the formula to work out the area of a circle:

a        = *(500/

a        = *(500/)*(500/)

a        = *(250000/²)

Simplified:

a        = 250000/

a        = 79577.47m²

The area of the circle is larger than the 30 side shape because a circle has infinite sides.

Conclusion

For any given n-sided shape, regular shapes have the maximum area and as the number of sides of a shape increase, the area increases.  A circle has the maximum area because it has infinite sides.  Therefore the farmer should fence off a circle with a perimeter 1000m because it would create an area of 79577.47m².

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

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