Some Isoperimetric Explorations,

Way back in my younger days I was a regional director for the Michigan Council of Math Teachers, and gave a lot of talks ... somewhere in that time I wrote some stuff about the type of optimization problems that mathematicians call isoperimetric problems.. Often they involved a given perimeter (a farmer has 360 feet of fence).. and the object is to find the maximum area under some given conditions (the area is a rectangle, etc..). Over the years I noticed some patterns to the types of solutions that often made the solution transparent with very little actual calculus or advanced math required. I wanted to write a couple of blogs to explain and, hopefully, organize my thoughts on them... but first, some introductory notes I wrote a long time ago about the problems for folks who are not familiar with the term isoperimetric, and to give a little history. The word isoperimetric literally means "same perimeter". It is usually used in mathematics to refer to to figures having the same perimeter but different shapes. Mathematicians use the term Isoperimetric Problem to describe problems relating to finding which of two figures with equal perimeter have the greatest area. The problems date back at least to Zenodorus , a 2nd century BC Greek mathematician who wrote On Isometric (same measure) Figures. In it he showed that of all isoperimetric polygons having the same number of sides, the regular polygon had the greatest area. [For example of all quadrilaterals having a perimeter of 16, the 4x4 square has the greatest area. (There is a hidden message in that statement that I will return to later.)] He also showed that the area of a circle was greater than any regular polygon with the same perimeter as the circumference as the circle. 

A related isoperimetric problem is called Dido's problem. Dido's problem is to find the maximum area for a figure with a given perimeter and bounded against a strait line. For example, if you had 100 feet of fence and wanted to enclose the maximum area with one side of the property along a straight river, or the side of a barn; what shape would enclose the maximum area. Dido was a Phoenician princess in Virgil's Epic tale, THE AENEID. The story tells of her founding of Carthage. When she fled from her brother, Pygmalion, who was trying to kill her (sibling problems seemed pretty dramatic in the Greek classics) she landed along the coast of Northern Africa and tried to buy land from the local king, King Jarbas. She was told she could have as much land as could be enclosed by a bull's hide. Dido had her followers cut the hide into tiny strips which were then strung together to make a great length, and enclosed a great area against the sea. That's where the strait line on one side comes from in the problem. Dido's solution was a semicircle. (pay attention here, this is one clue about generalizing the solution of isoperimetric problems when one side is against a natural straight boundary, stay tuned.) 

Dido's problem is sometime called the Problem of Hengist and Horsa for a similar conclusion to an English story about two wizards/lords who come from Germany to defend against a Saxon invasion in the fifth century. The two turn out to be the antagonists of the young King Arthur and Merlin. Even Dido's story is related to England in a round-about way. Her brother Pygmalion would create a beautiful statue and then fall in love with it. Aphrodite brought it to life for him. This transformation of a stone statue found its way through a George B. Shaw play into the famous English musical, My Fair Lady Perhaps one more small side-note to bring all this full circle to my interest in language, the model for Professor Henry Higgins in the play was a real live linguist named Henry Sweet who was instrumental in the development of the International Phonetic Alphabet in 1897. Ok, enough for now, so next I can examine one of the patterns I have noticed. Maybe you can find it too... here is a guided discovery tour... 1) A farmer has 360 feet of fencing, and wants to build two congruent rectangular pens that share a common fence. Maximize the total area (this is a common textbook problem... ) and find the dimensions. Now I have already mentioned the use of squares as the maximal area for isoperimetric quadrilaterals, but... well test your ideas.

2) step two is to do the same thing except one side of the area is bounded by a straight river.. so that side does not need to be fenced... 

3) step three is to extend the original problem to three congruent rectangular pens 

4) Now there are (at least) two ways to arrange four congruent pens, in a row or in a 2x2 square... try both ways... Now look at all the answers and pay close attention to the dimensions... I will return to this below,  so work out your ideas before you read on.  

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So we begin with two rectangular pens side by side.  Using the idea that squares are the isoperimetric ideal  among quadrilaterals, we find the size of each of the seven fence sections for congruent squares.  360 / 7 is 51 3/7, so the area of the two congruent squares would be \( A= 2(\frac{360}{7})^2 = 5289.7959...\)  [49ths have repeat periods of length 42, so forgive my truncating the answer].  

So what if we try some rectangles.  The total area will be h x w for the total height, and if we make h=w (squares again(sort of) the the height and width would each be 180 feet of fencing, with three "vertical" and two "horizontal"  sections,  That makes the 4 horizontal pieces each 45 units, and the three verticals each 60 units long.  Our Area = 2 x 45 x 60 = 5400 sq units.. and more than two squares.  

We can abstract the process by calling each horizontal piece of fencing h', and each vertical piece v', and we know that 3 v' + 4 h' =360.  and the total area is v' x 2h'.  Solving the first for v' gives us v' = (360 -4h')/3 , so the area is v' x 2h' = (360 -4h')/3 x 2(h') =(720h' +8h'^2)/3.  Since this is a quadratic, we know the max solution is halfway between the two roots, and so we can just ignore the divide by 3 since it won't change the roots.  One is obviously at zero,  720 + 8h' =0, and we get 90 units.... for the second zero , and 45 for h' at the max area .... but wait, looking back we see that is the solution we got in the last paragraph for equal fencing on the horizontal pieces and the vertical pieces.  

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So what about the two pens against a barn or  river?  (The student should convince himself that putting the river along the horizontal will give a larger total area than down a side.  If you can't, try both ways.  If I was wrong to use this approach, please drop me a note.) Now we have only two horizontal pieces and three vertical.  Proceeding as in the last model, we have 2h' + 3v' = 360, but the total area is still 2h' x v'.  now v'=(360 - 2h')/3, so we get an area in terms of h' as A= 2h' + (360 - 2h')/3 and disposing of the useless denominator again, we need to average the solutions of 720h'-4h'^2=0.  Ok h'=0 is easy, and simple division gives us h'=180 as the second zero.  So the maximum area occurs at h'=90 units,  That means the two horizontal pieces use up 180 units of fencing, and the three vertical sections divide the rest and are 60 units each. The total area is 180 x 60, or 10800 sq units.  This exactly doubles the area without the river by doubling the horizontal length. 

Note that there are no longer any squares in the rectangular pens, individually or in total.  You may, however, have picked up on a different "truth: suggested by these problems and the same problem for single rectangular pen....The total area is greatest when you use the same amount of horizontal fencing and vertical fencing (at least so far).

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So what happens when we go to three rectangular pens? And how shall we arrange our three pens?  Should we put three side by side, two above with a congruent one turned 90 degrees below, or will three squares closely nested, but not forming a rectangle be better,  We are rich with opportunities to experiment.   

Let's start with three squares, shaped like this Notice we can do the total fencing with five horizontal sections and five vertical sections, all of which are equal.  So each of ten equal fence pieces with a total fencing length of 360 must be 36 units each, or 3 squares with area of 36^2= 1296 sq units for a total fenced area of 3888 sq  units.  

So What would three pens all in a row give us.  We now have 4 vertical sides ans six horizontal sides.  Repeating the development of our beautiful quadratic solutions from before we solve from 6h' + 4v' = 360, and A = v' x 3h' we get 0=1080-18h' as the non zero root, or h=60 for the upper zero, and h=30 for the maximal area solution.  So the six horizontal edges use up exactly half the fencing, leaving 45 units for each vertical. so each rectangle is 30 x 45 for 1350 sq units in each of three rectangles to give a total fenced area of 4050 sq units.  So this exceeds the solution with three squares. But is this the best we could do?

What if the two squares on top were side by side with one congruent rectangle lying under them.  To make a rectangle of the whole configuration, the dimensions would have to have the long side= twice the width, as shown.  However you do this I think you will find that 14 of the widths will equal the total fencing, so each narrow length is 360/14 =180/7.  the way we drew the picture, the total area is two wide and three high, or total area = 3 x 180/7 x 2 x 180/7 = 6 x (180/7)^2 =3967.3469....  This is not a better solution, and it does not divide the fencing equally between the vertical and horizontal equally.  

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So let's explore four congruent pens in a rectangle.  If the four pens are all side by side in a row, we will have five vertical edges and eight horizontal.  The quadratic we need will have 5 v + 8 h = 360 .  (note I have dispensed with the primes.)  Our area relation is v x 4h = area.  So Area = (360 - 8h)/5 x 4h = (1440h-32h^2) /5.  So our largest root is at h=45, and our max area is when h=45/2.  Since there are eight of these fence pieces, they will use up 180 units, or half the available fencing; a good sign.  The five vertical sides of the pen will then be 36 units each, making each pen 810 sq units, for a total area of 3240 sq units. Is this the best possible for four pens?  

What if we tried a 2x2 square of Pens?  We have (in my drawing) four squares made with six vertical fences and six horizontals. certainly if we put 30 units on each of these sides, the horizontal and vertical fencing will be equal, but is our total area larger or smaller.  each pen will be 30^2 or 900 sq units, and the four in total will be 3600 sq units.  More than the four in a row.  

So equal use of fencing in both directions is, I'm believing, a necessary, but not sufficient for maximal solution.   You might think about what configurations you could use for five pens.  Obviously you could have five side by side, but by now you may be having doubts about that being the best for n>2.  You could stack four side by side over one rotated; or three side by side over two rotated. Are there more?  I can think of two more but not all congruent.  

I will return too this exploration soon, but want to add one more note about Fido-like problems.

Earlier I asked "If you have 360 feet of fencing and want to build a pen along a river or barn, ..".  The rectangular pen we found 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... " 

Of course you know at infinity you get a semicircle of fencing.  The sweet mathematical idea to make all these types of problems easy is symmetry . Imagine that you had found the perfect solution to the fence problem with three sections and used it on your side of the river. Then the guy on the other side of the river decided to copy your master plan on his side... NOW make the river really narrow..... no..more narrow... like a line.. Now erase the river completely... What do you see when you look at your neighbors fence and your fence together? It must be a six sided figure.. and it must contain the most area it could contain for any six sided figure, since both sides hold the maximum possible... you couldn't make the total bigger without making one side or the other bigger...so you must have the most efficient six sided figure possible, a regular hexagon. So what does your half look like? It must be half of a regular hexagon divided across two opposite vertices.

The solution would be to divide the 360 feet of fencing into three sections, and make the angles between the fence sections 120^o and slide the two ends up against the river. To solve the three sided non-symmetric figure, we used 1/2 of a six sided symmetric figure. 

The final area is three equilateral triangles with sides of 120 units.  I leave it to you to confirm that the area inside the semi-hexagon borders is almost double the rectangular river solution.  To give a visual of the difference, I've made a sketch of the two fenced areas with a scale of 60 units.  The river runs along the x-axis.  

And the general solution from there, I hope, is trivial. Actually with a little hand waving, this blog could have been much shorter..but I remember Pascal once wrote, "I have made this longer than usual, only because I have not had the time to make it shorter." [Lettres Provinciales, No. 16 (1657)]