Maths Fence Length Investigation
Extracts from this document...
Introduction
There is a need to make a fence that is 1000m long. The area inside the fence has to have the maximum area. I am investigating which shape would give this.



I am going to start investigating different shape rectangles, all which have a perimeter of 1000m. Below are 2 rectangles (not to scale) showing how different shapes with the same perimeter can have different areas.
350m
In a rectangle, any 2 different length sides will add up to 500, because each side has an opposite with the same length. Therefore in a rectangle of 100m X 400m, there are two sides opposite each other that are 100m long and 2 sides next to them that are opposite each other that are 400m long. This means that you can work out the area if you only have the length of one side. To work out the area of a rectangle with a bas length of 200m, I subtract 200 from 500, giving 300 and then times 200 by 300. I can put this into an equation form.
1000 = x(500 – x)
Below is a table of results, worked out using the formula. I have gone down by taking 10m off the base every time.
Height (m)  x  Area (m2) 
0  500  0 
10  490  4900 
20  480  9600 
30  470  14100 
40  460  18400 
50  450  22500 
60  440  26400 
70  430  30100 
80  420  33600 
90  410  36900 
100  400  40000 
110  390  42900 
120  380  45600 
130  370  48100 
140  360  50400 
150  350  52500 
160  340  54400 
170  330  56100 
180  320  57600 
190  310  58900 
200  300  60000 
210  290  60900 
220  280  61600 
230  270  62100 
240  260  62400 
250  250  62500 
260  240  62400 
270  230  62100 
280  220  61600 
290  210  60900 
300  200  60000 
310  190  58900 
320  180  57600 
330  170  56100 
340  160  54400 
350  150  52500 
360  140  50400 
370  130  48100 
380  120  45600 
390  110  42900 
400  100  40000 
410  90  36900 
420  80  33600 
430  70  30100 
440  60  26400 
450  50  22500 
460  40  18400 
470  30  14100 
480  20  9600 
490  10  4900 
500  0  0 
Using this formula I can draw a graph of base length against area.
As you can see, the graph has formed a parabola.
Middle
H2 = 150000
H = 387.298
½ X 200 X 387.298 = 38729.833m.
Below is a table of result for isosceles triangles from a base with 10m to a base with 500m.
Side = . Where X is the base length. I have used this formula to work out the area when the base is different heights. To do this I need to use Pythagoras’ theorem. Below is a diagram of the isosceles triangle.
Base (m)  Side (m)  Height (m)  Area (m ) 
0  500.0  500.000  0.000 
10  495.0  494.975  2474.874 
20  490.0  489.898  4898.979 
30  485.0  484.768  7271.520 
40  480.0  479.583  9591.663 
50  475.0  474.342  11858.541 
60  470.0  469.042  14071.247 
70  465.0  463.681  16228.832 
80  460.0  458.258  18330.303 
90  455.0  452.769  20374.617 
100  450.0  447.214  22360.680 
110  445.0  441.588  24287.342 
120  440.0  435.890  26153.394 
130  435.0  430.116  27957.557 
140  430.0  424.264  29698.485 
150  425.0  418.330  31374.751 
160  420.0  412.311  32984.845 
170  415.0  406.202  34527.163 
180  410.0  400.000  36000.000 
190  405.0  393.700  37401.537 
200  400.0  387.298  38729.833 
210  395.0  380.789  39982.809 
220  390.0  374.166  41158.231 
230  385.0  367.423  42253.698 
240  380.0  360.555  43266.615 
250  375.0  353.553  44194.174 
260  370.0  346.410  45033.321 
270  365.0  339.116  45780.727 
280  360.0  331.662  46432.747 
290  355.0  324.037  46985.370 
300  350.0  316.228  47434.165 
310  345.0  308.221  47774.209 
320  340.0  300.000  48000.000 
330  335.0  291.548  48105.353 
333.3  333.3  288.675  48112.522 
340  330.0  282.843  48083.261 
350  325.0  273.861  47925.724 
360  320.0  264.575  47623.524 
370  315.0  254.951  47165.931 
380  310.0  244.949  46540.305 
390  305.0  234.521  45731.554 
400  300.0  223.607  44721.360 
410  295.0  212.132  43487.067 
420  290.0  200.000  42000.000 
430  285.0  187.083  40222.817 
440  280.0  173.205  38105.118 
450  275.0  158.114  35575.624 
460  270.0  141.421  32526.912 
470  265.0  122.474  28781.504 
480  260.0  100.000  24000.000 
490  255.0  70.711  17324.116 
500  250.0  0.000  0.000 

Because the regular rectangle was the largest before, I added 333.
Conclusion
No. of sides  Area (m2) 
8  75444.174 
9  76318.817 
10  76942.088 
As you can see, the larger the number sides on the shape, the larger the area is. This pattern has carried on going for all of the shapes that I have investigated, so I am going to investigate shapes with the following amount of sides:
20, 50, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000.
Below is a table showing the results that I got.
No. of sides  Area (m2) 
20  78921.894 
50  79472.724 
100  79551.290 
200  79570.926 
500  79576.424 
1000  79577.210 
2000  79577.406 
5000  79577.461 
10000  79577.469 
20000  79577.471 
50000  79577.471 
100000  79577.471 
On the next page is a graph showing the No. of sides against the Area.
As you can see form the graphs, the line straightens out as the number of side’s increases. Because I am increasing the number of sides by large amounts and they are not changing I am going to see what the result is for a circle. Circles have an infinite amount of sides, so I cannot find the area by using the equation that I have used to find the other amount of sides out. I can find out the area of a circle by using o. To work out the circumference of a circle the equations is od. I can rearrange this so that the diameter is circumference/o. From that I can work out the area using the or2 equation.
1000/o = 318.310
318.310/2 = 159.155
o X 159.1552 = 79577.472m2
If I place this point on my graph it is at the same place, area wise, as the last results on my graph were. From this I conclude that a circle has the largest area when using a similar circumference
This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.
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