Fencing Problem
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
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 
410  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.
Middle

Side = . Where X is the base length. I have used this formula to work out the area when the base is different heights.
Conclusion
No. of sides  Area (m2) 
3  48112.522 
4  62500.000 
5  68819.096 
6  72168.784 
7  74161.644 
Now that I have this equation, I am going to use it to work out the area for a regular octagon, nonagon and decagon.
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 
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
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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