Math Fencing Project
I have to find the maximum area for a given perimeter (1000m) in this
project. I am going to start examining the rectangle because it is by
far the easiest shape to work with and is used lots in places (most
things use rectangles for design- basic cube .etc). To start with what
type of rectangle gives the best result.
A regular square or an irregular oblong?
I start by having 4 individual squares.
[IMAGE]
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[IMAGE][IMAGE] Goes to
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Regular square irregular oblong
Now look at how many sides are exposed on each shape-
å sides of each cube internal1 å sides of each cube internal2
[IMAGE][IMAGE]Ratio for square = ratio for oblong =
å sides of each cube exposed1 å sides of each cube exposed2
2 ´4 (1 ´ 2) + (2 ´ 2)
[IMAGE][IMAGE] = =
2 ´ 4 (3 ´ 2) + (2 ´ 2)
= 1 = 0.6
This can be further done by having more squares (to show that the more
irregular a square is the less area it has for that given perimeter.
BUT if we want the same perimeter (which we do) we have to take away a
square for the irregular oblong to make it the same area as the
regular square.
[IMAGE]
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Now look the irregular oblong has less area. So we've proved that for
rectangles. The more sides kept internal, the smaller the area. Now we
desimplify the length ´ width equation-
[IMAGE] ab
= ½ (a2 + b2) ´ ½ (a2 + b2) eventhough it would be easier to do ab,
this shows what I mean.
½ (a2 + b2) makes the sides even like a square
ab
[IMAGE] ½ (a2 + b2) times it by the ratio of its real area to a
squares. (like in percent)
or simply written
A = ab
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