The fencing problem 5-6 pages
The Fencing Problem - Mathematics A farmer has exactly 1000 meters of fencing, with it she wishes to fence off a plot of level land. She is not concerned about the shape of the plot, but it must have a perimeter of 1000m. She wishes to fence off a plot of land, which contains the maximum area. Part One Using rectangles; I will investigate how she may obtain a maximum plot of land, using rectangles with a perimeter of 1000m. I will do this by using calculations to find the area of each different possible perimeter situation. For the first part of the investigation, I will calculate some possible areas for the plot of land, making sure the perimeter is exactly 1000m. For my first examination, I will try quadrilaterals. Side 1: 100m Side 2: 400m Side 1: 200m Side 2: 300m Side 1: 250m Side 2: 250m Side 1: 450m Side2: 50m Side 1: 350m Side2: 150m Side 1 Side 2 Area 100 m 400 m 40000 m2 250 m 250 m 62500 m2 300 m 200 m 60000 m2 350 m 150 m 52500 m2 450 m 50 m 22500 m2 The graph and the table indicate that when the width and the height of the regular quadrilateral have the least difference, the Area seems to be at its greatest. The statement is proven by the square. A square has all the sides the same, so the difference is the smallest possible value since it is zero. The area of the square is the greatest, and this leads me to believe that regular shapes - i.e. shapes that have equal sides have the biggest areas. I will test this prediction out on triangles. I will start with the most regular triangle - the equilateral, and then test a number of isosceles triangles. If my prediction is true then the equilateral triangle will have the largest area. Side 1, 2,3= 333.333 Sin (60)= x/333.33 X=288.66 (height) Area= 1/2 X base X height =48112.52m2 Side 2, 3=450m Base= 100m 150m2 +x2= 350m2 x=316.23m (height) Area= 1/2 X base X height = 47494.16m2 Side 2, 3=400m Base= 200m 100m2 +x2=400m2 x=382.30 (height) Area=1/2 X base X height =38729.83m2 Side 2, 3=350m Base= 300m 150m2+x2=350m2 x=316.23m Area=1/2 X base X height =47434.16m2 Side 2, 3=300m Base=400m 200m2+x2=300m2 x=223.61m Area=1/2 X base X height =44721.36m2 Base Equal sides Area 100m 450m 22360.68m2 200m 400m 38729.83 m2 300m 350m 47434.16 m2 333.3m 333.3m 48112.52 m2 400m 300m 44721.36m2 This graph proves my initial theory. The equilateral triangle, which had the least difference between the sides, has the largest area. The equilateral triangle, which is the most regular of all the triangles since it has equal sides, has the biggest area. This is similar to the square, which is the most regular of all the quadrilaterals. Therefore I can now use only regular polygons (polygons with equal
sides) to gain the biggest areas. However I have also made another observation. The three-sided triangle has a smaller area than the four-sided square. It may just be possible that area increases with the number of sides. Consequently, this is my next hypothesis. To test my new theory, I will have to work out the area for other regular polygons, with a great number of sides. Part 2 ====== Investigate the shape, or shapes of the plot of land, which have the maximum area. I will start of with the pentagon. Each side= 200m (1000/5) All corresponding angles are the ...
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sides) to gain the biggest areas. However I have also made another observation. The three-sided triangle has a smaller area than the four-sided square. It may just be possible that area increases with the number of sides. Consequently, this is my next hypothesis. To test my new theory, I will have to work out the area for other regular polygons, with a great number of sides. Part 2 ====== Investigate the shape, or shapes of the plot of land, which have the maximum area. I will start of with the pentagon. Each side= 200m (1000/5) All corresponding angles are the same Exterior angle = 360/N or 360/5 = 72 Interior angles= 180-72=1080 (angles on a straight line) Like all regular shapes, a pentagon can be split up into equal triangles, which will be isosceles as two sides, and the corresponding angles are equal. To find the Area we use the formula for a triangle Area=1 X base X height. We know that the base is 200m since it is the same as one side. Nevertheless, we need to find the height of the triangle to find the Area. We know all the angles in the triangle so we can use trigonometry: Tan(72/2)= 100/H Rearranged this formula is: H=100/Tan46 H=137.64m Area=1 X base X height Area=0.5 X 200 X 127.64 Area = 13763.82m2 But this is only the Area of one of the triangle so we now have to multiply by 5. The Area of the Pentagon is=68819.10m2 If I look at the result of this Area, I see that my theory was correct, as the Area of a Pentagon is bigger than the Area of a square. Therefore, I can say that the Area of a regular shape increases as the shape has more sides. I can and will carry on testing this theory with other shapes such as hexagons, heptagons and octagons. However, I have also analyzed that the method used to find out the Area of a pentagon is quite lengthy, and it may be appropriate that I find out a formula that will aid me to quickly find out the Area of a regular shape at a later time. Now I will find the Area of a hexagon and see if there is any repetitive method that may help me find a formula. -Each side of a hexagon= 1000/n or 1000/6= 1662/3m -Exterior angle= 600 (360/6) -Interior angle = 1800-600=1200 Tan (300)=83.333/H H=144.34m Area=1 X base X height =1X144.34X1662/3 =12028.13m2 Multiply by 6 to find the Area of all the triangles=72168.78m2 From this I can say that the Area does increase with number of sides. The Area has continued rising, as the number of sides has increased, albeit at a slower rate- the difference between the Area of a hexagon and pentagon is not as great as the difference between the Area of a square and triangle. Now I can see the repetition in some aspects of the working out. I have found that I have been using the Area of a triangle as a foundation, since it has to be used to find the overall Area of a polygon. I have also used the Tan function to find the height of each triangle. So now, with my knowledge so far I can derive a formula to find the Area for an n sided regular polygon. I will start with the facts that I know about this shape: - Each of the sides is equal the perimeter is 1000. Therefore one side = 1000 /n All regular shapes can be split up into equal triangles. The number of sides will determine the number of triangles that can fit into the shape - the n sided shape has n amount of equal triangles in it. Area of a triangle=1 X base X height Exterior angle of triangle = 360/n Interior of triangle = 180 - 360/n these facts have enabled me to derive a formula. Here is how I did it: - For one triangle inside the n sided shape- the Area is: 1 X base X height - For an n amount of triangles the Area is: n(1 X base X height) However, in the formula - n(1 X base X height)- what is the base and what is the height? The base of the triangle = the length of one side Therefore the base = 1000/n n(1 X 1000/n X height) Now we have to find the height of the triangle. Previously when finding the height we have had to use trigonometry before using the Area equation. Therefore, to find the height we will need to know an angle and a length. We can place a perpendicular bisector through the triangle, which creates a right angle. Now we can use the Tan function to find the height. We know that the base of the triangle is 1000/n. Therefore, the angle is 360/n, which is the same as the exterior angle. However, to find out the height of the triangle we only need half of this angle: 1 X 360/n = 360/2n or simplified = 180/n We can use the Tan function to find out the height: Tan (180/n)= Opposite/Adjacent or 1X 1000/n/height Rearranged: Height= 1X 1000/n/ Tan (180/n) Our new Area formula is: A= n ({1 X 1000/n} X 1X 1000/n/ Tan (180/n)) We can simplify this Area formula by multiplying out brackets to: = n( 500/n X [500/n / Tan (180/n)] ) = n( 500/n X 500 /( n X Tan (180/n)) ) = n( 250000/n2 X Tan (180/n) ) = 250000/n X Tan (180/n) This is the formula for an n sided shape with a 1000m perimeter. If I now work out a series of regular shapes, starting at an octagon, it may be possible to find out what shape will have the largest Area. Octagon Area = 250000/n X Tan (180/n) = 250000/8 X Tan (180/8) = 75444.17 m2 Nonagon Area = 250000/n X Tan (180/n) = 250000/9 X Tan (180/9) = 76318.82 m2 Decagon Area = 250000/n X Tan (180/n) = 250000/10 X Tan (180/10) = 74161.48 m2 11-sided shape Area = 250000/n X Tan (180/n) = 250000/11 X Tan (180/11) = 77401.98 m2 12-sided shape Area = 250000/n X Tan (180/n) = 250000/12 X Tan (180/12) = 77751.06 m2 13-sided shape Area = 250000/n X Tan (180/n) = 250000/13 X Tan (180/13) = 78022.50m2 20-sided shape Area = 250000/n X Tan (180/n) = 250000/20 X Tan (180/20) = 78921.89393m2 50-sided shape Area = 250000/n X Tan (180/n) = 250000/50 X Tan (180/50) = 79472.72422m2 100-sided shape Area = 250000/n X Tan (180/n) = 250000/100 X Tan (180/100) =79551.28988m2 I can see that the area is still rising albeit at a slower rate each time. There is clearly not much of a difference between the 50 and the 100-sided shape. There are more faster ways of working out a series of equations- I.e. spreadsheets. However, before I do that, I shall work out one more area. If the area increases with the number of sides, the greatest area will be the shape with the greatest number of sides. The biggest number is infinity, so the shape with infinite sides has the biggest area. A circle is the only shape with infinite sides. As the number of sides on a shape gets bigger, each of the sides gets smaller, and closer together. In the circle, the sides have become small dots that join up in a round fashion. Since the number of sides on a circle is infinite, you cannot use the equation. The letter n would become obsolete. Therefore, you have to work out the area of a circle a different way. Again I will have to use the information I know to work out the area of a circle: Perimeter/circumference = 2 x � x Radius Perimeter= 1000m Area= � X Radius2 Therefore: 1000m=2 x � x Radius Rearranged that equals: Radius= 1000m/2 x � =159.15m Consequently, we can work out: Area of circle= � X Radius2 =� x (159.15) 2 =79577.47m2 The area of circle is clearly the greatest so far, and will be the greatest of all. I now have a range of results that will allow me to plot results on a graph and show how the area changes as the number of sides increase. These results are very interesting. The spreadsheet clearly shows that the rate that the area increases and slows down dramatically, which tells me that there is not much difference in making a fence with 36 sides and 37 sides. The point for the area of a circle would be much to the right of this graph. Although there is no halt in the increase, the area only increases by a few hundredths at a time. The maximum point would be the circle, so the graph would never exceed 79577.47m2 in area. The results spreadsheet is more useful though. I had to split the equation up to easily get a result. You can clearly see how the area column increases very slowly after a while. However, what is interesting is that the equations denominator creeps towards the number PI --- 3.141593. If the equation n X Tan (180/n) is tending towards PI then when the denominator actually reaches Pi the equation must be at its biggest point, i.e. the equation must equal the area of a circle. The area column also is very close to the area of a circle and continually gets bigger. In summary this would mean that the area of the circle with 1000m circumferences is 250,000/ � that is a much quicker way of working out the area. Conclusion ---------- After looking at many different shapes there are a few basic conclusions that I have made: Regular shapes i.e. shapes that have equal sides, have bigger areas than irregular shapes. This is proved by the quadrilaterals and the triangles the more sides a regular shape has, the bigger area it has. (Pentagon, hexagon etc.) I have also learnt that the area for a regular shape with 1000m perimeter can be worked with a simple formula. To cover any other extension work, I could work out the formula for any regular shape by changing my formula like thus: If I take my original formula in its least simple form I could alter my equation to deal with all perimeters: A= n ({1 X 1000/n} X 1X 1000/n/ Tan (180/n)) Since the Perimeter in this formula is 1000, I can replace 1000 with p (perimeter) each time. The new formula is: A= n ({1 X p/n} X 1X p/n/ Tan (180/n)) I can simplify this equation now: A= n ({1 X p/n} X 1X p/n/ Tan (180/n)) A= n (p/2n) X p/2n / Tan (180/n)) A= n (p/2n) X p/2n / Tan (180/n)) A= n(P2/ 4n2 x Tan(180/n)) A=P2/ 4n x Tan(180/n) This is the formula for all regular shapes. The only area that this and my original formula will get unstuck is for circles. I have found out that a circle has infinite sides so when substituted into the equation: A=250000/ n x Tan (180/n) A=250000/ ∞ x Tan (180/∞) This is not a workable equation. However, after more investigation, I found out that the denominator of theoriginal equation is equal to Pi when n is a very large number. Therefore, the area of a Triangle with a 1000m perimeter is: A=250000/� However, it is possible to work out the area of a circle with any perimeter as a piece of extension work, if you devise another formula. If you use the formula for above for 1000m circular perimeter, and convert it to the formula to find thearea for any regular shape, you will get this formula for the circle. A=P2/4� If we go back to the original question, we can see that the farmer wants to find thebiggest area for her fencing. Thebiggest area has been shown as the circle in this investigation, as it has the greatest amount of sides possible, which is infinity. After using graphs and tables, and any previous knowledge, I have proved that the circle has thegreatest area and the farmer should make a circular enclosure with her 1000m of fencing. This circle will have a radius of: 1000/ 2 x � = 159.15m And an area of: A= 250,000/� or =79577.47m2 which is larger than any other shape shown in this investigation.
