Math Fencing Project

1125 Words3 Pages

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]

[IMAGE]

[IMAGE]

[IMAGE]

[IMAGE][IMAGE] Goes to

[IMAGE]

[IMAGE]

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]

[IMAGE]

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

Open Document