Volume $V=\pi r^2 h$.
Surface area $$S =2 \pi r h +2 \pi r^2$$
Now, the AM-GM inequality says
$$\frac{ \pi r h + \pi rh +2 \pi r^2}{3} \geq \sqrt[3]{\pi r h \cdot \pi rh \cdot2 \pi r^2}= \sqrt[3]{2 \pi V^2 }$$
with equality if and only if $ \pi r h = \pi rh =2 \pi r^2$.
Basically, it boils down to this simple principle (which is more or less AM-GM): if some expressions have a constant product, their sums is minimal if the functions are equal (if they can be equal).
We are apparently in a 2D world here, so a “box” is in fact a rectangle, which has an area and a perimeter. The most trivial answer to “why aren't the two always proportional” would be a counter-question of “why should they”.
But this is all about intuition, so imagine you start with a cardboard square of $10\times10$ units. It has a perimeter of $4\times 10=40$ units. Cut it in half and you obtain two rectangles of $10\times 5$ units, each of which has $30$ units perimeter. That's because your newly introduced cut created new perimeter in what used to be the inside, namely twice the length of the cut, i.e. $2\times10=20$ units. OK, two rectangles are not one rectangle, but if you glue them together along their short edge, you loose $2\times5=10$ units of perimeter which now are glued and therefore contained in the inside, but you still have $2\times(5+20)=50$ units of perimeter; more than what you started with. So in this example, the cardboard is the “stuff” whose amount stays fixed, and the perimeter depends on the shape, can be gained by cutting and lost by gluing.
But there is the opposite case as well. Instead of a cardboard square, take a piece of string, $40$ units of length. If you form a square, you will get the same square as above, covering an area of $100$ square units. But if you create other figures with the string, you get other areas. So here the perimeter is what stays fixed, and the area varies.
There are a lot of topics related to this relation between perimeter and area:
- Of all rectangles with given area, the square has least perimeter. Conversely, of all rectangles with given perimeter, the square has largest area.
- Of all shapes with given area, the circle has least perimeter. Conversely…
- For a given area, the perimeter can become arbitrary large. One can even define objects with finite area but infinite perimeter, like e.g. the fractal Koch snowflake.
- Measuring perimeter in real life can depend a lot on scale. The classical question here is how long is the coast of Britain? (Which is also the name of a paper by Mandelbrot.)
Actually, on a side note, how IS the perimeter measured, is it thought of as part of the content? Though if not, how is it measured, what is measured?
The perimeter of a planar figure is a simple case of a topological boundary and as such not part of the “content” (i.e. interior) of the shape. As I just mentioned, measuring a boundaries in real life using a ruler can lead to results which depend on the ruler. So in most branches of mathematics, the problem is more one of computing the perimeter. One common approach starts by finding a piecewise parametric description of the perimeter, then integrating over that. But it depends a lot on what kind of shape you are talking about, how it is described.
What's also fascinating is that this doesn't seem possible with circles, yet with any other object it seems that it is.
A circle has a fixed ratio between area and perimeter. So does the square. The class of all rectangles has no such ratio, but if you e.g. fix the aspect ratio of the rectangles, e.g. only consider 16:9 TV screens, then you again have a fixed ratio between area and perimeter. If you want a more flexible class containing circles, you could consider all ellipses. There again the ratio between area and perimeter depends on the actual shape (and computing the perimeter can be really hard). So it all depends on what family of shapes you consider, whether knowing the area fixes the perimeter (or vice versa) or not.
Best Answer
Tie the two ends of a thread several inches long together. Fill your kitchen sink with water. Let the waves settle down. Let the loop of thread float on the placid surface, but place it so that the thread meanders erratically --- it's not a circle but a meandering curve returning to its starting point. Then drop a tiny drop of liquid dish detergent into the part of the surface surrounded by the thread. The detergent is a surfactant: it decreases the surface tension. But it does so in the region surrounded by the thread and not outside it. The result: surface tension on the outside pulls the thread outward and it suddenly assumes a circular shape!
Conclusion: the circle surrounds a larger area than does any other closed curve of the same length.
So think about that.
That's the "physicists' solution" of the isoperimetric problem. Dym & McKean's book on Fourier transforms solves the same problem by using Fourier series.