# How can an object grow so that the ratio of surface area to volume remains constant?

Obviously there are many applications for this and many solutions. I am also interested in closed curves that have the same ratio of area to arc length as we "grow" them, growth is done by augmentation in both cases.

-
This is not possible: the ratio of surface area to volume is not "dimensionless". If you scale the object by λ, the ratio will grow accordingly. – Fabian May 9 '11 at 17:29
@Fabian That just means that you cannot scale the solutions to this question. The question makes sense. For example, if you think of animals, their volume tells you how much heat they generate and their surface tells you how much heat they lose. So individual animals and species cannot grow by scaling. (Well, at least not for long without adapting a lot of other things.) – Phira May 9 '11 at 17:35
@user9325: I was reading "growth is done by augmentation" in terms of scaling. Of course, if you don't scale them then it can be done. But is the question then to characterize all curves who have the ratio 1 (in some units) of surface to volume? – Fabian May 9 '11 at 17:39
I doubt there is anything that can be said about all such curves, I would just like a few elementary... Not too ugly examples... If such a thing exists. – futurebird May 9 '11 at 17:44
@Fabian I think that the problem is to formulate the right question. – Phira May 9 '11 at 17:44

An example: Start out with a square of size 1 and now grow one side as $x$ and one side as $y(x)$, then you want $(x+y)=2x y$ or $y(x)=\frac{x}{2x-1}$.

So after some time, one side stays constant $1/2$ and the object grows "only in one direction".

I doubt that you can get much more precise answers without having a more precise question, but it is possible that someone else knows the right question.

-
You did "beat me" by 12 seconds ;-) – Fabian May 9 '11 at 17:45
This works, if we think of the augmented bit as an upside down "L" shape. – futurebird May 9 '11 at 18:23

Take a rectangle (with sides $a$ and $b$). The "volume" is given by $V=a b$; the "surface" is given by $S=2(a+b)$. Fix the ratio to $$r = \frac{S}{V} = \frac{2(a+b)}{ab}.$$ You can solve this equation for $$b = \frac{2a}{a r -2},$$ valid for $ar>2$. Plugging it into the expression for the volume, we obtain $$V= \frac{2 a^2}{a r -2}.$$ You can see that the volume grows arbitrarily large as $a\to2/r$.

-