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Fencing Problem In this piece of coursework I shall try to obtain the biggest possible area enclosed by 1800m of fencing. I shall look at different types of sizes and shapes of fences to obtain the biggest area. First I shall look at rectangles as they are easiest to calculate. P=2L+2w 1800=2L+2w 900=L+w 900-w=L I shall use this formula to work out different lengths by substituting a different number for w. A=L x w A=(900-w) x w w l=900-w Area=w(900-w) 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900-50=850 900-100=800 900-150=750 900-200=700 900-250=650 900-300=600 900-350=550 900-400=500 900-450=450 900-500=400 900-550=350 900-600=300 900-650=250 900-700=200 900-750=150 900-800=100 900-850=50 50(900-50)=42500 100(900-100)=80000 150(900-150)=112500 200(900-200)=140000 250(900-250)=162500 300(900-300)=180000 350(900-350)=192500 400(900-400)=200000 450(900-450)=202500 500(900-500)=200000 550(900-550)=192500 600(900-600)=180000 650(900-650)=162500 700(900-700)=140000 750(900-750)=112500 800(900-800)=80000 850(900-850)=42500 I tabulated the results because it allows you to see the different areas with the different lengths. This table shows that a rectangle of both lengths and widths of 450 (square) gives the biggest area in terms of the rectangle. I shall plot my results onto a graph to see the correlation between them. The line of symmetry shows the biggest area when you have 1800m of fencing in the shape of a rectangle. I found the ideal dimensions were if the length and width...

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