The Fencing Problem

A furthermer has exactly g-force metres of fencing with it she wishes to fence off a plot of primer. She is not concerned astir(predicate) the shape of the plot, but it essential see a border of c0m. So it could be or anything else with a circumference (or circumference) of curtilagem What she does wish to do is fence off the plot of land which contains the level best playing reach. Investigate the shape, or shapes of the plot of land which have a upper limit champaign.Throughout this investigation I pass oning check that the perimeter is 1000 meters by reining the total of each(prenominal) the outer sides. also I allow use refining as a bureau of decision the maximum country. When I talk about utilise the maximum eye socket of the previous tabularise the maximum celestial orbit of each circumvent go forth be highlighted.Rect locomotesThe first shape I leave al wizard audition impart be a rectangle. Having been told that the perimeter essential be 10 00 meters I allow rally the scopes of three rectangles, each with different continuances of sides, fashioning sure that the perimeter is kept the akin.To calculate the athletic field I result use the order LENGTH x WIDTH = AREAor playing field = lw.Rectangle A l = 450mw = 10mArea = 450 x 10Area = 4500m2Rectangle B l = 300mw = two hundredmArea = 300 x 200Area = 60000m2Rectangle C l = 100mw = 400mArea = 100 x 400Area = 40000m2Having carried out the to a higher(prenominal) place calculations I allow for create a spreadsheet with facial expressi peerless to carry out much calculations. The trends go away populate of Length, comprehensiveness, edge and Area. Under length in that respect result be a variable number (less than 500 and greater than 0). The first convention will be put under the width read/write head. The width will be calculated by taking the length away from 500. This will undertake the perimeter to be 1000m.The linguistic rule will be =500-B2 whe re B2 is the cell in which the length is. To double check that the perimeter is 1000m under the perimeter heading at that place will be another formula. This will be =(B2+C2)*2 where B2 is the length and C2 is the perimeter. It will be multiplied by 2 because the resolve in the brackets would be just the total of ii sides and not all four. last under the area heading there will be a formula. This will be =B2*C2 where B2 is the length and C2 is the width. This formula is the same as the hotshot apply previously to calculate the area of a rectangle. The formulas and headings will be gained in as shown in the table infra.Length (m)Width (m)Perimeter (m)Area ( satisfying m)490=500-B2=(B2+C2)*2=B2*C2Having entered the correct information I will be able to calculate the areas of many different sizes of rectangles with a perimeter of 1000m. I can do this in Microsoft surmount by force the formula boxes down, thus duplicating them but allowing them to refer to different lengths.(Pl ease master tables and represents Fencing business for Rectangles)To start with I used my spreadsheet to contract the area of a some rectangles within the throw of 1m and 499m.I accordingly plotted a graph show length against area. It showed a holy curve. I intractable that the line of unanimity of this curve would help to find the length that would ruin me the maximum area. I institute the line of symmetry to be along the 250m mark on the x axis of the graph. schemeI predict that the length of a rectangle that will give me the maximum area will be 250m. I have decided this having embed the line of symmetry on the graph. queen regnant (Please see tables and graphs Fencing bother for Rectangles)To prove my hypothesis I refined my look for approximately the maximum area of the first table and then the sulfur table, followed by the third table and so on. Eventually I found that, even to 1 decimal place above or below 250m that, the maximum area was given by rectangle of sides 250m by 250m. This shows that a upstanding up gives the maximum area for a rectangle. symmetric TrianglesThe second shape that I will try will be an symmetric trilateral. Having carried out essays for a rectangle I am going to see whether the maximum area will be bigger, smaller or the same as that of a rectangle. I am also going to find out whether the number of sides affects the results and whether there are any similarities in results to a trilateral. This will help me find the shape that gives the maximum area.As previously for rectangles I will test some different sized isosceles triplicitys that have an area of 1000m.The formula for the area of a triangle is BASE x HEIGHT shared by 2 or bh/2. I cannot find the area without well-read what the heyday of the triangle is. To find the visor of the triangle I must use Pythagoras. This states that for a right-angled triangle a2+b2=c2 or the form hypotenuse is oppose to the sum of unbent of the other two side s. Therefore to find the visor I must split the triangle in half and then use half of the ancestor to help me find the height. The square height will therefore be rival to the square of the hypotenuse minus the square of half the base. In the below examplesb = base, s = one touch on side of the triangle and h = height.Triangle A b = 500ms = 250mb/2 = 250mh = 2502-2502h = 0mArea = 250 x 0 / 2Area = 0m2Triangle B b = 400ms = 300mb/2 = 200mh = 3002-1002h = ?50000mh = 223.6068mArea = 400 x 223.6068 / 2Area = 44721.35955m2Triangle C b = 200ms = 400mb/2 = 100mh = 4002-1002h = ?150000mh = 387.29833mArea = 200 x 387.29833 / 2Area = 38729.38466m2After completing the above tests I will create a spreadsheet with formulae to carry out more calculations. The headings will incorporate of low, 1 play off 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 i sosceles 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 equate calculations carried out. These are put at the front of the formula and the substitute for square al-Qaida is SQRT. So my formula will be the square root of 1 equal side squared, minus half the base squared. yet before entering my formula I found out that using the power sign () doesnt give accurate results and in order to square metrical composition I must multiply the number by itself preferably of using such a sign. Therefore the formula entered into the spreadsheet will be=SQRT((C2*C2)-((B2/2)*(B2/2)))Finally under the ar ea heading there will be a formula. This will be =(B2*E2)/2 where B2 is the base and E2 is the height. This formula is the same as the one used previously to calculate the area of a triangle. The formulas and headings will be entered in as shown in the table below.Base (m)1 Equal Side (m)Perimeter (m)Height (m)Area (square m)200=(1000-B2)/2=B2+(C2*2)=SQRT((C2*C2)-((B2/2)*(B2/2)))=(B2*E2)/2Having entered the correct information I will be able to calculate the areas of many different sizes of isosceles triangles with a perimeter of 1000m. I can do this in Microsoft Excel by drag the formula boxes down, thus duplicating them but allowing them to refer to a different base.(Please see tables and graphs Fencing chore for Isosceles Triangles)As before I entered a range bases between 1m and 499m. I then plotted a graph of base against area and found that unlike the results for a rectangle there wasnt a perfect curve in order to find the line of symmetry, to aid my search. notwithstanding I could tell that the maximum area would be given by a triangle with a base between 300m and 400mHypothesisI predict that the maximum area will be given by a triangle with equal sides. I have decided this because the maximum area for a rectangle was given by a square and that my graph shows that the base must be between 300m and 400m. For a triangle with equal sides and a perimeter of 1000m the base would be 333.33meters.Poof (Please see tables Fencing Problem for Isosceles Triangles)To prove my hypothesis I refined my search around the maximum area of the first table and then the second table, followed by the third table and so on. Eventually I found that, to 2 decimal places, the maximum area was given by a triangle of equal sides which is 333.33m to every side. This shows that an equilateral triangle gives the maximum area for a triangle and this proves my hypothesis right.Regular PolygonsHaving tested isosceles triangles and rectangles I found that unconstipated sided shapes gi ve the maximum area. I know this because the maximum area of an isosceles triangle is given when the sides are each 333.33m. The maximum area given by a rectangle is give by a square with 250m sides. I have also that as you gain the number of sides the area increases because the maximum area for a rectangle is 62500m2, and the maximum area for an isosceles triangle is 48112.52243m2. As a result of these findings I am going to test perpetual sided polygons.Having split the pentagon into isosceles triangles and then into right angled triangles I can now find the area. I know that the base of the triangle is 100m however I do not know the height. Before finding the height I must work out what the inhering angle is. To find this I will divide 360 by the number of right-angled triangles (in this case 10). I can now tell the following about the triangle I can now use Trigonometry to find the height of the triangle.SOH CAH TOAI know what the opposite is and the angle, and I penury to k now what the adjacent is. I will therefore use the formula TAN=Opposite/Adjacent. Therefore Adjacent=Opposite/TAN. So the height in metres will beHeight = 100/TAN36Height = 137.638192mArea of 1 Isosceles Triangle = (200*137.638192)/2Area of 1 Isosceles Triangle = 13763.819205m2Area of Pentagon = 13763.819205*5Area of Pentagon = 68819.09602 m2After completing the above tests I will create a spreadsheet with formulae to carry out more calculations. The headings will consist of Number of Sides, 1 Equal Side, Perimeter, Internal wobble of 1 Triangle, half(a) Angle, Height (of internal isosceles triangle), Area of 1 Triangle and Total Area. Under the first heading (Number of Sides) there will be a variable, whole, number between 3 and as higher number as desired (e.g. 30). Under the second heading there will be a formula to calculate the length of one equal side. The formula will be =1000/A3 where A3 is the number of sides. As in all the other tests there will be a formula to check tha t the perimeter is 1000m. This will tell me if I have made an error in any of the previous cells.So far so good, however before I continue I must point out that a computer spreadsheet doesnt work in degrees to nib angles. It measures in radians where a release rotation is 2?. Also ? is represented by PI() in a spreadsheet. So instead of using 360 in my formula under the Internal Angle of 1 Triangle heading I will use 2*PI()/A3 where A3 is the number of sides. Under the Half Angle heading there will be a formula that will be =D3/2 where D3 is the internal angle of one triangle. This gives the internal angle of 1 right-angled triangle.My main formula will go under the Height heading and it will use Tan which is substituted by TAN in a spreadsheet. It will be =(B3/2)/TAN(E3) where B3 is 1 equal side and E3 is the angle inside a right-angled triangle. The area of one isosceles triangle will be calculated using the formula =(B3*F3)/2 where B3 is one equal side and F3 is the height. Fi nally the total area will be calculated by multiplying the area of one isosceles triangle by the number of sides. The formula entered will be =G3*A3 where G3 is the area of one triangle and A3 is the number of sides. The formulas and headings will be entered in as shown in the table below.Number1 Equal SidePerimeterInternal AngleHalf AngleHeightArea of 1 TriangleTotal Areaof Sides(m)(m)of 1 Triangle (rad.)(rad.)(m)(square m)(square m)5=1000/A3=B3*A3=2*PI()/A3=D3/2=(B3/2)/TAN(E3)=(B3*F3)/2=G3*A3Having entered the correct information I will be able to calculate the areas of many uniform polygons with different numbers of sides and with a perimeter of 1000m. I can do this in Microsoft Excel by dragging the formula boxes down, thus duplicating them but allowing them to refer to a different number of sides.HypothesisI predict that as you increase the number of sides the area increases because the maximum area for a rectangle is 62500m2, and the maximum area for an isosceles triangle is 48112.52243m2. conclusion (Please see graph and table Fencing Problem for Regular Polygons)Used my spreadsheet to calculate the areas of polygons with sides ranging from 3 to 30. The polygons with 3 and 4 sides were used to test that my formula worked correctly. I plotted a graph cover the number of sides against the area and found that, as predicted, as the number of sides change magnitude so too did the area.CircleAfter my findings from carrying out tests on regular polygons I have decided to test circle. I have decided this because as the number of sides of a regular polygon increase so too does the area and a circle is an infinitely sided regular polygon.HypothesisI 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 2?r and ?r2. The circumference must be 1000m and before finding the area I need to find the radius. rung = (1000/2)/?r = 500/?r = 159.1549431mArea = ?*159.15494312Area = 79577.47155m2To 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 SidesCircumference (m)Radius (m)Area(square m)Infinite1000=(C2/2)/PI()=PI()*(D2*D2)Formula2?r(Circumference/2)/??r2ProofNumber of SidesCircumference (m)Radius (m)Area(square m)Infinite1000159.154943179577.47155The table above clearly proves my hypothesis correct. The working out also proves my hypothesis correct.ConclusionHaving completed the spreadsheet table I can conclude that a circle gives the maximum area and th at 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