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
800=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
00
50
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
00(900-100)=80000
50(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 of the rectangle were 450m. This gives you the biggest area of 202500cm².
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
800=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
00
50
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
00(900-100)=80000
50(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 of the rectangle were 450m. This gives you the biggest area of 202500cm².