Friday, 19 November 2010

Repeat of an Isoperimetric Discovery

It's that time of year again, so I want to repeat this to explain some ideas I laid out in Calculus today...

Recently we have been doing maxima/minima applications in calculus. By the time they get her most of them know that when they maximize the area of a rectangle with perimeter constraints, they get a square. In Pre-calc we do some basic problems with 2 adjacent pens, a single pen against a wall or river, and all the similar cases. I have written about the generalizations of some of these here, and here. The problem of two adjacent congruent pens is (or should be) easy for a calculus or precalc student. It turns out that the maximum occurs when the total width is one-fourth the perimeter and the height is one-sixth the perimeter, for a maximum area of one twenty-fourth of the perimeter value (in square units, of course). If you extend that to even more congruent rectangular pens placed end to end it turns out that the width stays the same?

If you extend this, as many teachers do, to a shape like this one, it starts to get interesting to the more observant student. The height is now one-eighth of the total , and the width is one-sixth of the perimeter. Even slow students can figure out that the area of the total turns out to be one forty-eighth of the value of the perimeter. Some will even generalize the area of an array of mxn congruent pens to be the perimeter squared divided by [4(m+1)(n+1))].... but it seems to take forever for some one to finally realize that in all these problems, whether there was one rectangle, or one with one side adjacent to a river or barn, or rows of rectangles... there is one invariant...

Theorem one for rectangular arrays of congruent rectangular pens "The amount of fence going right to left is the same as the amount going up and down." If you want to maximize a rectangular array with five rows of three pens in each row, we will need 24 lines of fence, four going vertically, six going horizontally. Use half the fencing to make the four vertical fences, and the other half to make the other six going horizontally. (Prove..... dear reader) ..

I always thought this fifty-fifty use of fencing was a cute mathematical "teacher trick" but I have recently come to believe there is something deeper at work. I played around recently with this problem but loosened the constraints on the pens to be just equal area rectangles rather than congruent. This offered some nice areas to explore. Which would be better, three congruent pens in a row, or two pens with a third beneath... like so. The three adjacent pens turn out to have a max area when the width was p/4 (where p is the perimeter) and the height was p/8, giving a maximum area of p2/32... and as before, that mean that half the perimeter was used for the vertical sections, and half for the horizontal sections..and for the three non-congruent sections with equal area the width was p/6, and the length was 3P/16, giving a max area of....wait, that's also p2/32. (Ok, I hadn't expected that)... but when I checked the dimensions, it turns out that the horizontal sections use up exactly half the perimeter, and the vertical sections the other half in both cases.... hmmm...

Ok, but every kid knows that the area is maximized when the rectangle is closest to a square... so what if we just put three squares arranged as shown below and didn't require that the whole assembly forms a rectangle... I thought that would probably be even more area.. It turns out that it isn't, The equal distribution of left and right still works out, but the total area is only 3p2/100 or about 96% of the other shapes. (ok, I wasn't ready for that)...

So, I decided to move up to five pens... I passed over four absolutely sure that the division into four equal squares in a two by two array would be the most efficient, but five offered several variations, five in a row; two rows of two with a single in the third row, or three in one row and two in another... The five congruent squares in a row gave a max area of p2/48, but the set up with three pens in one row and two in another gave a slightly larger area of 5p2/216 or p2/43.2. The 2,2,1 set up (shown below) gave not quite the same area.. This set up gave a max area of p2/44.8.

From all of this I have a formed a totally unproven conjecture about the maximum area when n equal area rectangular pens are set out in a rectangular boundary with a given amount of fencing.
The Theorem One above for congruent rectangles seems to be a stronger property than I suspected, and is applicable to all equal area pens in a rectangular boundary, And the conjecture is that when the number of pens, n, is between x2-x and x2+x, the maximum area will be found when there are x-rows of pens and all of the pens have either x-1, x, or x+1 pens in the row.

The number of pens in any row will never differ by more than one from any other row. For values of n that are perfect squares or pronic (product of consecutive integers), the number of pens in each row will be the same. When n > x2 there will n-x2 rows that have x+1 pens, and the remaining rows will all have x pens. Similarly, if n is less than x2 there will be x2 - n rows that have x-1 pens and the remaining rows will have x pens each. As an example, if n= 23, then x=5 since 52-5 is less than 23 which is less than 52+5. And for 23 pens there will be five rows with x=five pens in three of them, and x-1=four pens in two of them.

I also have discovered that the theorem about equal lengths of fencing horizontally and vertically seems to be more important as a "rule of thumb" for maximizing than is equal sides. For example if five squares are set out as efficiently as possible (four in a square with one attached to a side), the area will be p2 / 45 with 8/15 of the perimeter going one way(say horizontally) and 7/15 going the other. If we change the squares to rectangles with each having a height of p/16 and a width of p/14 the area is p2/44.8.. an improvement of about 1/2 %....
Even if you go to six pens (where there is no lost perimeter enclosing the odd pen), putting two rows of three squares gave me an area of P2 / (48 1/3) but using the equal division of fencing gives the very slightly improved maximum of p2/48.

I can't explain why, but the equal division of fencing theorem is stronger than I first suspected, and if someone out there can shed more light on it than my feeble exploration, please do.

I haven't had time to expand the formula for the max area of all n when a rectangle is divided into congruent area rectangles, (not too difficult, I think).. but for square or pronic numbers of pens, it seems the max area will always be p2/(4 (x+1)(y+1)) where xy=n. As a limit as n approaches infinity, it seems the max area for rectangles divided into n equal area rectangles approaches
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