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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

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Introduction

Amir Taaki        Maths 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.

image00.pngimage00.pngimage00.pngimage00.png

image00.pngimage00.pngimage00.pngimage00.png

                                Goes toimage00.pngimage00.png

image00.pngimage00.png

                                                 Regular square                   irregular oblong

Now look at how many sides are exposed on each shape-

∑ sides of each cube internal1                                                       ∑ sides of each cube internal2

Ratio for square =                                                      ratio for oblong  =image02.pngimage03.png

∑ sides of each cube exposed1                                   ∑ sides of each cube exposed2

                               2 × 4                                                                    (1 × 2) + (2 × 2)

             =                                                          =image05.pngimage04.png

                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.

image00.pngimage00.png

image00.pngimage00.pngimage00.png

image00.pngimage00.png

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-

                              abimage06.png

= ½ (a2 + b2) ×  ½ (a2

...read more.

Middle

image12.png

image12.pngimage12.pngimage12.pngimage12.pngimage12.pngimage12.png

Left one has less perimeter.

Because more sides are kept internal. Well this can be used on right angled triangles.

image13.pngimage13.pngimage13.png

image14.png

image12.pngimage12.pngimage14.png

image15.pngimage15.pngimage15.png

                                                 s

image14.png

image12.pngimage12.pngimage14.png

                                                           r                                

Getting the idea?

But the more r becomes different from s, the more those squares inside the triangle become disproportional. That’s why the best ratio for r and s would be 1 : 1.

Also

This has more circumference                            than this. this is because you always cover more of a And less area                                                     route in less Steps by going from A to B

image19.pngimage16.pngimage18.pngimage17.png

image17.pngimage19.png

image17.pngimage19.png

image07.png

                                                (this can apply to everything which means I don’t

have to finish or prove anything more but I’ll do it the hard way)

This is because this

image20.pngimage21.pngimage22.png

                       Leaves that triangle empty. And guaranteed that will be less than 2s because of

                         Pythagoras theoreum

image23.pngimage22.pngimage24.png

(Now think about it if a square isn’t as much area when disproportioned sides- and a square is made up of triangles, what will it be when you have a triangle).

This then applies to equilateral triangles being better than all other triangles because a)

image39.png

(which is why squares have more area than triangles,

...read more.

Conclusion

We know its regular as it stays constant from a point.

image40.png

or in another way.

----------------------------------------------------------------------------------------------------------------------

* a = side of square

   s = radius

   c = perimeter

   A = Area

               A    =   π× (s × s)

                      = 3.141… × s2

              A     =   a2  =   3.141… × s2

Same areas

But!

               C    =   π× 2s

                      = 6.283185… × s

              C     =   (π× 2) ×s × 4   =   4 ( 2πs

               =  10.0265131 ×s

Now that difference in 4 is nothing compared to square root which shrink huge numbers down to tiny sizes. The smaller a number you square root, the less that is taken away (bear in mind that s > 0). But we are dealing with 1000, here, not 2, so the cut would be significant.

Therefore we can say that circles should be considered multi sided instead of 1 sided.

Also we can say the more sides a shape has the bigger its area for a given circumference, also sectors of a shape being of the same shape also makes the area bigger (sectors being the center of a shape joined to all corners- including circle).

What is this area then that we get with a circle?

R = 1000 / 2π

R = 159.1549431…

A = π× (159.1549431…)2

A = 79577.47155

Lets just back check against a square.

S = 1000 / 4

S = 250

A = 2502

A = 62500

Circle is biggest. No need for number crunching to prove anything, but if you want it there are spreadsheets Behind here is spreadsheets.

Username: alex                 Created on 28/03/2003 10:48 AM

...read more.

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