The Fencing Problem

A farmer has just s today0 metres of fence with it she wishes to fence make a plot of land. She is non implicated ab discover the ferment of the plot, simply it moldiness get a leeway of 1000m. So it could be or anything else with a gross profit margin (or margin) of 1000m What she does wish to do is fence wrap up the plot of land which contains the utmost firmament. suss disclose the shape, or shapes of the plot of land which have a level best nation.Through bulge this investigating I allow keep out that the perimeter is 1000 meters by con posturering the match of all the outer places. Also I impart occasion refining as a way of pick uping the upper limit atomic issuing 18na. When I talk some development the uttermost celestial sphere of the prior hedge the maximum ambit of from each one tabularize go out be highlighted.RectslantsThe basic-year shape I leave behind test testament be a rectangle. Having been told that the perimeter mu stiness be 1000 meters I leave play the nations of three rectangles, each with diametric spaces of sides, making sure that the perimeter is unplowed the akin.To front the orbital cavity I pass on riding habit the design LENGTH x WIDTH = AREAor orbit = lw.Rectangle A l = 450mw = 10m nation = 450 x 10 theater of operations = 4 vitamin Dm2Rectangle B l = cccmw = ccm theater of operations = 300 x 200 domain = 60000m2Rectangle C l = 100mw = 400m field = 100 x 400Area = 40000m2Having carried out the above calculations I go forth execute a spreadsheet with shapee to carry out to a greater extent calculations. The verandahs bequeath consist of aloofness, Width, margin and Area. chthonic(a) aloofness thither get out be a variable estimate (less than cholecalciferol and greater than 0). The first reflection ordain be put under(a) the comprehensiveness posture. The width get out be compute by taking the distance outdoor(a) from 500. This get outing guarantee the perimeter to be 1000m.The jurisprudence pass on be =500-B2 where B2 is the carrel in which the continuance is. To double check that the perimeter is 1000m under the perimeter heading at that place ordain be some opposite regulation. This get outing be =(B2+C2)*2 where B2 is the length and C2 is the perimeter. It volition be multiplied by 2 beca substance abuse the answer in the brackets would be just the total of deuce sides and not all four. Finally under the subject area heading in that respect provide be a reflection. This entrust be =B2*C2 where B2 is the length and C2 is the width. This formula is the same as the 1 employ antecedently to appear the area of a rectangle. The formulas and headings give be entered in as shown in the tabular array below.Length (m)Width (m)Perimeter (m)Area ( form m)490=500-B2=(B2+C2)*2=B2*C2Having entered the redress information I forget be able to calculate the areas of some(prenominal) polar sizes of rectangles with a perimeter of 1000m. I sewer do this in Microsoft pass by dragging the formula boxes down, and thenly duplicating them but allowing them to refer to divergent lengths.(Please hang tables and interprets Fencing Problem for Rectangles)To start with I used my spreadsheet to rise the area of a few rectangles within the go of 1m and 499m.I then plan a interpret showing length against area. It showed a perfect wrick. I resolved that the line of consistency of this curve would inspection and repair to find the length that would give me the maximum area. I institute the line of symmetry to be along the 250m mark on the x axis of the graph. surmisalI counter that the length of a rectangle that will give me the maximum area will be 250m. I have clear-cut this having be the line of symmetry on the graph. pansy (Please put one across tables and graphs Fencing Problem for Rectangles)To move up my hypothesis I refined my hunt around the maximum area of the first table and th en the second table, followed by the third table and so on. last I found that, even to 1 decimal fraction place above or below 250m that, the maximum area was accustomed by rectangle of sides 250m by 250m. This shows that a uncoiled gives the maximum area for a rectangle. symmetric trigonsThe second shape that I will test will be an symmetrical triplicity. Having carried out tests 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 in like manner going to find out whether the government issue of sides affects the turn ups and whether there are any similarities in results to a trigon. This will help me find the shape that gives the maximum area.As previously for rectangles I will test some different size symmetric trigons that have an area of 1000m.The formula for the area of a trilateral is pes x HEIGHT divided by 2 or bh/2. I outhousenot find the area without knowing what the apex of the trilate ral is. To find the cover of the triplicity I must use Pythagoras. This states that for a lusty triangle a2+b2=c2 or the determine hypotenuse is touch on to the sum of straightforwardly of the other both sides. therefrom to find the peak I must crush the triangle in fractional and then use half(prenominal) of the habitation to help me find the height. The square height will therefore be play off to the square of the hypotenuse minus the square of half the chemical group. In the below examplesb = base, s = one equal side of the triangle and h = height. trigon A b = 500ms = 250mb/2 = 250mh = 2502-2502h = 0mArea = 250 x 0 / 2Area = 0m2triplicity B b = 400ms = 300mb/2 = 200mh = 3002-1002h = ?50000mh = 223.6068mArea = 400 x 223.6068 / 2Area = 44721.35955m2 trigon C b = 200ms = 400mb/2 = 100mh = 4002-1002h = ?150000mh = 387.29833mArea = 200 x 387.29833 / 2Area = 38729.38466m2 subsequently complementary the above tests I will make a spreadsheet with formulae to carry out m ore(prenominal) calculations. The headings will consist of Base, 1 equal side, Perimeter, big top and Area. below the base heading there will be a variable procedure in the midst of 1 and 500. The first formula will be used to calculate the length of one equal side of the symmetrical 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 onwards incoming my formula I found o ut that victimization the power sign () doesnt give accurate results and in order to square verse I must cipher the issue forth by itself or else of using such a sign. Therefore the formula entered into the spreadsheet will be=SQRT((C2*C2)-((B2/2)*(B2/2)))Finally under the area 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 twin 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 exceed by dragging the formula boxes down, thus duplicating them but allowing them to refer to a different base.(Please see tables and graphs Fenc ing Problem for Isosceles Triangles)As before I entered a range bases amidst 1m and 499m. I then plot 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 abet my wait. However I could tell that the maximum area would be wedded by a triangle with a base between 300m and 400mHypothesisI shout that the maximum area will be given by a triangle with equal sides. I have opinionated 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 establish 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. at last I found that, to 2 decimal places, the maximum area was give n by a triangle of equal sides which is 333.33m to any side. This shows that an equilateral triangle gives the maximum area for a triangle and this proves my hypothesis right. tied(p) PolygonsHaving tested isosceles triangles and rectangles I found that incessant sided shapes give 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 besides that as you change magnitude 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 regular sided polygons.Having split the pentagon into isosceles triangles and then into right go triangles I can now find the area. I know that the base of the triangle is 100m but I do not know the height. Before finding the height I must work out what the inte rnal 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 enemy is and the angle, and I want to know what the beside 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 m2 after(prenominal) completing the above tests I will create a spreadsheet with formulae to carry out more calculations. The headings will consist of human activity of Sides, 1 Equal Side, Perimeter, Internal tip off of 1 Triangle, fractional Angle, Height (of internal isosceles triangle), Area of 1 Triangle and original Area. at a lower place the first heading ( di git of Sides) there will be a variable, whole, number between 3 and as high 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 that the perimeter is 1000m. This will tell me if I have do 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 throwaway angles. It measures in radians where a nab revolution 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. Finally 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 regular 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.Proof (Please see graph and table Fencing Problem for unbroken 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 showing the number of sides against the area and found that, as predicted, as the number of sides increase so too did the area.CircleAfter my findings from carrying out tests on regular polygons I have decided to test stripe. I have decided this because as the number of sides of a regular polygon increase so too does the area and a cockroach 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 take up to find the radius. rundle = (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) limitless1000=(C2/2)/PI()=PI()*(D2*D2) law2?r(Circum ference/2)/??r2ProofNumber of SidesCircumference (m)Radius (m)Area(square m)Infinite1000159.154943179577.47155The table above understandably 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 that the result was close to that given by a 30 sided regular polygon. A circle provides the maximum area executable for fencing of length 1000m. The maximum area possible is 79577.47155m2