Saturday, 24 January 2009

Patterns in Iso-perimetric Problems-2

In my first post on iso-perimetric problems, I ended with some questions, all related to, " A farmer has 360 feet of fencing, and wants to build " where in each case the farmer wanted to build a number of rectangular pens, and in one case no fence was needed along one edge as that was contained by a "straight river" (this is often replaced in textbook problems by making the pens against a barn or house).

In the first case, for two congruent pens, the maximal area I found was when each pen was 45 ft. along the sides that are not shared, and 60 ft in the direction dividing the two pens. for a total area of the two pens of 5400 sq. ft.

In the second case, if the river or barn replaces the fence on one side of two of the adjacent pens, the maximal area occurs when the sides parallel to the river are each 90 feet (twice as big as the last problem) and the remaining three sections perpendicular to this are each still 60 feet. There is a clue to a generalization in the fact that the new solution moves all the fence that would have been used on the river side of the pens to the opposite side and none to the ends.

In the case where there are three adjacent pens in a row, the solution is for each of the pens to be 30 feet along the sides that runs along all three pens, and 45 feet in the direction that divdes the pens.

Is there anything we could observe about these solutions that might answer the two remaining cases with four pens? "Yes!" he answers emphatically. Imagine each of the problems above with fences running north-south and east-west. In each solution, the amount of fencing used in the north-south direction is exactly equal to the amount used in the east-west direction, 180 feet, or one-half of the 360 feet available.

If we assume that will work in the two cases with four pens, then the one with four adjacent pens will have five fence lengths in the direction dividing the line of pens, and two long sections across both ends of the four pens... 180/8 gives 22.5 feet, and 180/5 = 36 feet, so each of the four pens will be 22.5 by 36 feet, and the total area of all the pens is 36 x 90 = 3240 square feet.
In the case of four pens arrainged in a square (don't you just know the four smaller pens will all be squares?), there are six fence sections along the east-west sides, and six along the north-south sides, so each fence was 30x30= 900 square feet, for a total of 360 square feet. A slightly better use of fencing than the four in a row...

In general, and that is where we are going... to maximize rectangular pens with M rows of N pens, and a total perimeter of P then the area of each pen will be P/2 divided by M(N+1) and in the other direction P/2 divided by N(M+1) so that half the fencing is used in each direction.

And as a foot note, as we noted in the case of four pens, 2pens x 2 pens is better than 4x1, and if we have R pens, we want to divide R into two factors of R that are closest together.

OK, so Onward to another relationship... If you have 360 feet of fencing and want to build a pen along a river or barn, the rectangular pen above is NOT the most efficient way to use three straight sections... assume you are limited to four fence posts, so you put two somewhere on the river, and can put two more along the fencing to hold straight sections of fence; how do you place them?

Now extend that problem to four sections of fence (and five fence posts)... and then generalize the results... Any takers?

More later... but I think to give you time in between, my next note will be about history, and in particular the reason that you don't know about George Joachim of Rhaetia.... admit it, you are excited already..

1 comment:

Anonymous said...

WOW, I've used problems like these in my calc and pre-calc classe. I never noticed that...Thanks