Area and volume?

Start with a 2D square with a side length of Z. What volume can be made from its surface area? Im not asking for a specific answer, partly because after some experimenting i see that this question is not explicitly answerable in this form.

I realized there are several ways to get several volumes and it all depends on the approach one takes

My first instinct:

divide Z^2 into 9 equal areas and use 5/9th's of this area to create a five sided cube. then take the remaining 4/9th's of the area and divide those 9ths and so on and so forth.

the end result for this surface area changed to volume is

n=0

4^n(Z^3/3^3)+ 4^(n+1)(Z^3/3^6)+ 4^(n+2)(Z^3/3^9)+...........

and I was fine with that until morning when i realized Z could also = 2*Pi*r and the volume = Pi *( Z/2*Pi)^2 * (2*Pi(Z/2*Pi))

after doing some numerical testing, the values i got were a good bit different.

then in hopes of finding a volume that is closer to my right circular cylinder value i decided to use 4/9 of Z^2 to make my cube
which would give me a volume of

2(Z/3)^3 + 2(Z/9)^3 + 2(Z/27)^3 + ..........

this produced a volume that was closer to my cirrcular cylinder , but not quite?

so how is volume measured? is in in little cubes or not?

are my methods accurate? what is the true max Volume?😕

oh and if at all possible keep this simple so i can undestand...thanks

Originally posted by joe shmo
Start with a 2D square with a side length of Z. What volume can be made from its surface area? Im not asking for a specific answer, partly because after some experimenting i see that this question is not explicitly answerable in this form.

I realized there are several ways to get several volumes and it all depends on the approach one takes

My first ...[text shortened]... e true max Volume?😕

oh and if at all possible keep this simple so i can undestand...thanks

Are you asking which form should a volume have to get the minimal surface area to volume ratio?
That would be a sphere (I think)
As for your little squares, wouldn't you need 6/9 of the little squares to make a closed surface?

Originally posted by serigado
Are you asking which form should a volume have to get the minimal surface area to volume ratio?
That would be a sphere (I think)
As for your little squares, wouldn't you need 6/9 of the little squares to make a closed surface?

Ok so the spherical surface area brings the most volume....I had another person mention that to me when discussing the problem

As for the six surface area's need to make a cube, not sure

the way i look at it there has to be a minimum of 4/9 and a maximum of 6/9

that is what is strange to me for defining volume.....how can 3 different surface areas contain the same volume?

with 4 you have an open top and bottom cube

with 5 you have an opended top or bottom cube

with 6 there is an entirley closed cube

this all is strange to me....do not all of the gemotrical objects have the same volume?

Originally posted by joe shmo
Ok so the spherical surface area brings the most volume....I had another person mention that to me when discussing the problem

As for the six surface area's need to make a cube, not sure

the way i look at it there has to be a minimum of 4/9 and a maximum of 6/9

that is what is strange to me for defining volume.....how can 3 different surface areas ...[text shortened]... cube

this all is strange to me....do not all of the gemotrical objects have the same volume?

try to think of volume's relationship to surface area in 3D as similar to the relationship between area and perimeter in 2D

you can clearly make closed geometric objects with different values for area that have the same perimeter. for example, take a 4x4 square (perimeter 16, area 16) and a 2x6 rectangle (perimeter 16, area 12). this includes non-polygon shapes: shapes with curved sides, etc. and it turns out in 2D (a planar world) that the shape with the greatest area/perimeter ratio is indeed the circle.

the analogy in 3D is where we look at the volume (how much space is enclosed) and surface area of a CLOSED 3D object. if you allow for open-top boxes, or even open-top open-BOTTOM boxes or something of the sort, you are skewing your result. i.e. volume stays the same in that the theoretical side you removed is still being considered as present, however your surface area is diminishing quite a lot - hence your volume/surface area ratio increases. but what's to stop you from removing all six sides and having a theoretical object with volume but no surface area?

lastly, as others have said before, the optimal shape for the greatest space enclosed within a minimal surface area (so, for instance, given 100 sq. feet of wrapping paper what's the biggest volume i could wrap up?) is indeed a sphere. 🙂

Originally posted by Aetherael
lastly, as others have said before, the optimal shape for the greatest space enclosed within a minimal surface area (so, for instance, given 100 sq. feet of wrapping paper what's the biggest volume i could wrap up?) is indeed a sphere. 🙂

Though a rectangular shape would be much easier to wrap 🙂

Originally posted by Aetherael
try to think of volume's relationship to surface area in 3D as similar to the relationship between area and perimeter in 2D

you can clearly make closed geometric objects with different values for area that have the same perimeter. for example, take a 4x4 square (perimeter 16, area 16) and a 2x6 rectangle (perimeter 16, area 12). this includes non-po ...[text shortened]... sq. feet of wrapping paper what's the biggest volume i could wrap up?) is indeed a sphere. 🙂

That is a very good question..

It seemed like 4 sides for my cube was the minimum, simply because I had never used a 3 sided cube in any application..there are open boxes and closed boxes and boxes without tops, so thats why i drew the line at 4

altough now I agree, what is really stopping me from taking another side off? its kind of a cool question, and that is exactly what i would like to know.

Are there applications involving objects with volume and without surface area?

So if Z is still the length of our square the the max volume it could contain with the least suface area is

V = 4/3*Pi*(Z/2*sqrt(Pi))^3

??

Originally posted by joe shmo
That is a very good question..

It seemed like 4 sides for my cube was the minimum, simply because I had never used a 3 sided cube in any application..there are open boxes and closed boxes and boxes without tops, so thats why i drew the line at 4

altough now I agree, what is really stopping me from taking another side off? its kind of a cool question, ...[text shortened]... e max volume it could contain with the least suface area is

V = 4/3*Pi*(Z/2*sqrt(Pi))^3

??

or am i over simplifiying that transformation