Saturday, 28 March 2009

Isoperimetric problems and scale

We had just finished that old classic about finding the minimum surface area of a can (closed cylinder) if the volume was fixed. I had given them a volume of 128 Pi cubic cm so that when they were done the radius and height would come out in integer units, four and eight cm respectively. Many found the minimum surface area to be 96 Pi square units and the ones who finished quickly were helping the others find their way through the substitution of h=128/r2 into the formula for surface area so that they could graph the function, or perhaps work out intricate details of arithmetic, "but 256 divided by 4 is 64, not 84." Finally we were ready, I thought, for a little closure.

"So, what did you notice?" Many hands up.... I point, they answer.. "The minimum is 96." I hesitated, perhaps even made a slight face of despair. I had hoped they would notice more. They picked up on the "That's not what I'm looking for" face and someone quickly added, "96 cubic centimeters... no meters, no, wait, I mean square centimeters.." amidst some laughter on their part I tried to draw them deeper into the problem. "Well, yes, that is the minimum for THIS problem, but what else did you notice?"

With long troubled faces, they poured through their solutions, what was the old man after now... we got the answer... finally a brave soul offers a trial balloon..."The surface area is less than the volume?"; drawn out as a question.
We had recently talked about the isoperimetric problems, planer figures with the same number for the area and perimeter, and I had talked about, and solved the cube with the same number for its volume and surface area... (I have several blogs a while ago on this idea; here and here and a third one here) which I assume made him think to compare the results in this fashion.
"So... " I asked. They somehow took that as encouragement and someone conjectured that the very act of "minimizing" the surface area meant it would be smaller than the volume. That seemed easy enough to counter, but I still wanted them to notice one other thing, so I thought a minute about which way to adjust and gave a second version of the same problem..."Find the minimum surface area of a cylinder with a volume of 16 pi cubic cm."
They sped quickly through the now-familiar problem, and announced that the minimum area was 24 Pi cubic inches. I waited, eyebrows lifted. A moment later the quickest realizes.... audibly..."Damn," and then to the student on his left, "It's bigger... its more than the volume!".

"So what ELSE do you notice, what seems to be true in both cases?" I ask. They struggle, nothing seems obviously the same. Someone notices, and offers, almost apologetically, that both the radius and height in the second problem were 1/2 of original solution .

I was willing to drag this out to make them think, so I asked her if she thought the radius and height would continue to diminish by 1/2 in every problem I could think up. Finally she restated what she thought might be a truism..."The radius is half the height, is that always true?"

"Well, let's test it." Earlier in the year they would have groaned when their answers led to another problem, but by now they expected that every question led to another question. This time I picked an intermediate value, knowing I only had one more chance to offer integer values, I suggested they try 54 pi cubic cm. They sped through the problem and found that indeed the height was 6, twice the radius value of 3. They were so glad to feel like they were on the track of something they didn't even notice that this time the surface area also had a numeric value of 54 pi.

I had allowed myself to get hung up between two big ideas, now I had to kill one to chase the other, so I just summarized their radius is one-half height in a simple statement. "Ok, so what have we found? The can which most efficiently contains a given volume of substance will have a height that is 2r, or equal to its diameter. It is the same distance across as it is up and down." I have always intended, and never followed up on, to make a solid that meets this equality. Maybe now I will; because the next thing I told them was wrong. I said that it was close to a typical tuna fish can. (actually the ones in my cupboard are too short for the diameter. The closest I could find was a can of Eagle Brand that was about 3.25 inches high and about 3 inches in diameter.)

Finally I wanted to try once more to drive home a big idea about that classic relationship between length, area, and volume that seems forever to not seem obvious to even advanced students. I talk about how in one case the volume was more than the surface area, and in another the volume was less, and then, we stumble upon the "Goldilocks" size, where they both are "just right."

I spend some time drawing rectangles of every kind of size and showing that you can always make one with the same ratio of sides that has exactly the same number for area and perimeter, and the same thing is true for any two or three dimensional shape... and then I said it, "Every shape is isoperimetric under some measurement scale!" and the kid sitting in the front row, before he even thought, says "That can't be so?, there must be some shapes that won't work."

So I set them one more problem; "I have a solid shape so convoluted that it has no name It has a volume of 64, and a surface area of 512 units. Blowing it up and shrinking it is like changing from measuring in inches to feet or mm or something different so let's just change the numbers. If I make a scale model of it that is bigger or smaller, what happens to the volume and surface area?"

They want a formula... I refuse...I return to the original two cylinder problems and ask what is the relationship between the original radii... they finally recall that it was four cm in the first and 2 cm in the last, the ratio was 2:1. "What about the heights?" and again they come up with 2:1...

"Ok, what about the surface areas?" I get 96/24 a few times and someone realizes 4:1...
"The volumes????" and they start to recall... Oh yeah, 8:1 squares and cubes...

"So if the volume is 64 and the surface area is 512, what happens if I make the model half as big in any length?" They realize that heads them off the wrong way and begin to work on the solution..

Finally one comes to the board with a solution... The ratio of SA : Volume = 8:1. If we increase the size by a linear dilation of k, the new ratio will be k2 8 : k3 and we want that to equal one. so we just set k2 8 = k3 which means when k=8 the ratio and surface area are the same number. "And the original shape???" I ask... "Doesn't matter." he answers... and mercifully for all of us, the bell rings.