0
$\begingroup$

I just found out from my calculations the following:

  • Corresponding a given length of a one-dimensional element, if a two-dimensional lamina, (having same boundary length as the length of the one-dimensional element) is constructed, then a circle has the largest area.

  • Corresponding a given area of a two-dimensional lamina, if a three-dimensional structure,(having same surface-area as the area of the two-dimensional lamina) is constructed, then a sphere has the largest volume.

Now let us call length, width, height as "parameters" of the first, second, and third dimensions.

Then, I state:

  • Any 1-D element has only one parameter-length.
  • Any 2-D lamina has two parameters-length, width. Product of these gives Area of lamina.
  • Any 3-D structure has three parameters-length, width, height. Product of these gives Volume of structure.

  • Thus, mathematically, when constructing an "n"-dimensional entity, from an "n-1"-dimensional entity, the product of all n-dimensions is the greatest for that entity, that consists all of its "boundary-points", in n-dimensional space, equidistant from the origin of the n-dimensional coordinates-system.

I guess that this (the last statement) could be proved (or rather, disproved) using principles of mathematical-induction, but that's just a mere guess.

Please help.

$\endgroup$
  • $\begingroup$ Could you tell a bit how did you calculated that the circle has the maximum area for a given parameter? $\endgroup$ – user103816 Jun 13 '14 at 11:09
  • $\begingroup$ Crossposted from physics.stackexchange.com/q/119091/2451 $\endgroup$ – Qmechanic Jun 13 '14 at 12:35
  • $\begingroup$ LOL, looks like someone from Phys.SE downvoted your question here. $\endgroup$ – user103816 Jun 13 '14 at 13:08
  • 1
    $\begingroup$ @user31782: I guess I forgot to mention my intuition. I just carried out for pairs of circle-square, and sphere-cube. I had come across while studying Euclidean-geometry that for a given perimeter, the circle possesses the largest-area. With a bit of intuition, I dared to proceed regarding sphere-cube, and from there, to n-dimensional Euclidean space. I guess that's why there's lack of precision in my question, as user111187 has pointed out. $\endgroup$ – Sudeepan Datta Jun 14 '14 at 16:25
0
$\begingroup$

This result (reformulated in a proper way, but it is what you mean) is known as (a generalization of) the isoperimetric inequality. For example, see this webpage where it is proved using calculus of variations (not induction).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.