This site has been archived for historical purposes. These pages are no longer being updated.
The honeycomb is a study in efficiency
Introduction: The art of optimizationThe mathematician's pursuit of extremes in his the persistent search to find the most honey using the smallest comb, the greatest profit for least expense, the greatest volume for least area, the greatest area for least perimeter, the most efficient universe, the smallest surface, the strongest beam, the least resistance, the shortest distance, the greatest capacity, the best estimate, the most useful shape, the comeliest form.
Why are soap bubbles always spherical?
The shortest distance between two points in a plane and on a sphere, a circle enclosing the largest area, and a sphere enclosing the largest volume were all known to the Greeks, even though the results were often stated without a real attempt at a proof. Another Greek discoveries, ascribed to Heron of the first century AD concerned a light ray taking the shortest path from point P to a mirror and back to point Q. This discovery that can be considered the germ of the theory of geometrical optics.
If one considers an arbitrary closed curve in the plane of a fixed perimeter, it is the circle that encloses the largest area. The ancient Greeks were aware of this, and many civilizations since then have intuitively used the shape of a circle, when they built their round cities, homes and castles. "Isoperimetrics", which examines the largest area surrounded by a fixed perimeter perhaps started with the "Dido problem". It is told of Dido, sister of Pygmalion and the daughter of King Mutton of Tyre in Phoenicia around the eighth century BC fled to the northern tip of Africa. There she purchased "as much land as she could surround with the hide of an ox." Dido cut the hide into a long string and then chose the shape of a semicircle as the largest area she could acquire (one border was on the Mediterranean Sea), and this settlement later became the thriving city of Carthage.
It is only natural that mathematicians should be interested in questions of this sort. In daily life problems of maxima and minima, of the "best" and the "worst," arise constantly. Many problems of practical importance present themselves in this form. For example, how should a boat be shaped so as to have the least possible resistance in water? What cylindrical container made from a given amount of material has a maximum volume?
Pierre de Fermat (1601 1665)Starting in the seventeenth century, the general theory of extreme values‹maxima and minima‹has become one of the systematic integrating principles of science. Fermat's first steps in his differential calculus were prompted by the desire to study questions of maxima and minima by general methods. In the century that followed, the scope of these methods was greatly widened by the invention of the "calculus of variations." It became increasingly apparent that the physical laws of nature are most adequately expressed in terms of a minimum principle that provides a natural access to a more or less complete solution of particular problems.
Certain branches of math specialize in extremes, but here I have divided the material into methods based on geometry, calculus and calculus of variations. I have taken this material from various sources, but especially from AN INTRODUCTION TO THE HISTORY OF MATHEMATICS by Howard Eves, THE CALCULUS by Otto Toeplitz, AN INTRODUCTION TO THE CALCULUS OF VARIATION by Charles Fox, MATHEMATICAL THOUGHT FROM ANCIENT TO MODERN TIMES by Morris Kline and WHAT IS MATHEMATICS? by Richard Courant. All sources are listed in the bibliography below.
Examples of superlatives using arguments from geometry
1. Maximum Area of a Triangle (see figure #1) (given two sides)
Given two line segments a and b, it is required to find the triangle of maximum area having a and b as sides. The solution is simply the right triangle whose two legs are a and b. If h is the altitude on the base a, then the area of the triangle is A = 2ah. Now 2ah is clearly a maximum when h is largest, and this occurs when h coincides with b; that is, for a right triangle. Hence the maximum area is 2ab.
2. Heron's Theorem (see figure #2) : Least distance property during the reflection of light Rays
Given a line L and two points P and Q on the same side of L. For what point R on L is PR + RQ the shortest path from P to L to Q? This is Heron's problem of the light ray.
If L were the bank of a stream, and someone had to go from P to Q as fast as possible, fetching a pail of water from L on the way, then he would have to solve just this problem.
To find the solution, we reflect P in L as in a mirror, obtaining the point P' such that L is the perpendicular bisector of PP'. The line P'Q intersects L in the required point R. It is simple to prove that PR + RQ is smaller than PR' + R'Q for any other point R' on L. For PR = P'R and PR' = P'R'; hence, PR + RQ = PER + RQ = P'Q and PR' + R'Q = PER' + R'Q. But PER' + R'Q is greater than P'Q (since the sum of any two sides of a triangle is greater than the third side), hence PR' + R'Q is greater than PR + RQ, which was to be proved.
Examples 1 to 6
3. Applications of Heron's theorem a minimum perimeter and a maximum area
a) Triangle of minimum perimeter (with given base and area).
Given the area A and one side c = PQ of a triangle; among all such triangles to determine the one for which the sum of the other sides a and b is smallest.
Prescribing the side c and the area A of a triangle is equivalent to prescribing the side c and the altitude h on c, since A = hc/2. The problem is therefore to find a point R such that the distance from R to the line PQ is equal to the given a, and such that the sum a + b is a minimum. From the first condition it follows that R must lie on the line parallel to PQ at a distance h. The answer is given by Heron's theorem for the special case where P and Q are equally distant from L: the required triangle PRQ is isosceles.
b) Triangle of maximum area (see figure #3)
In a triangle let one side c and the sum a + b of the two other sides be given; to find among all such triangles the one with the largest area. The solution is the isosceles triangle for which a = b. This triangle has the minimum value of a + b for its area; that is, any other triangle with the base c and the same area has a greater value of a + b. Moreover, it is clear from that any triangle with base c and an area greater than that of the isosceles triangle also has a greater value of a + b. Hence any other triangle with the same values of a + b and of c must have a smaller area, so that the isosceles triangle provides the maximum area for given c and a + b.
4. The extremum property found in thelocus of an ellipse
The problem of Heron is connected with some important geometrical theorems. If R is the point on L such that PR + RQ is a minimum, then PR and RQ make equal angles with L. This minimum total distance we shall call 2a. Let p and q denote the distances from any point in the plane to P and Q respectively, and consider the locus of all points in the plane for which p + q = 2a. This locus is an ellipse, with P and Q as foci, that passes through the point R on the line L. Moreover, L must be tangent to the ellipse at R. If L intersected the ellipse at a point other than R. there would be a segment of L lying inside the ellipse
5. Schwarz's triangle with least perimeter (see figure #4)
By using the method of reflection, Hermann Schwarz (1843-1921) solved the least perimeter of a triangle problem:
How to inscribe an acute-angled triangle in another triangle with the least possible perimeter. The answer is: The triangle vertices are the three points where the altitudes of the given triangle meet the opposite sides.
6 Steiner's problem (see figure #5)
Jacob Steiner solved the problem of the three villages A,B,C which are to be joined by a system of roads of minimum total length: i.e. three points A, B, C are given in a plane, and a fourth point P in the plane is sought so that the sum a + b + c shall be a minimum, where a, b, c denote the three distances from P to A, B, C respectively.
The answer to this problem is the following. If in the triangle ABC all angles are less than 120°, then P is the point from which each of the three sides, AB, BC, CA, subtends an angle of 120°. If, however, an angle of ABC, e.g. the angle at C, is equal to or larger than 120°, then the point P coincides with the vertex C.
7 The Golden rectangle is the comeliest of all rectangles
The partheon has the shape of a golden rectangle
The golden rectangleWith their sense of beauty and proportion, the Greeks came to regard certain shapes as more pleasing than others and to build many of their buildings in those shapes. The most famous example of this is what is called the golden ratio, the ratio of the lengths of the sides of a rectangle with the most pleasing proportion. Suppose that AB/BC = AC/AB. In this case, point B is said to divide AC in extreme and mean ratio, and, if the length of BC is 1, the length of AB is the golden ratio,
f = (1+5)1/2/2 = 1.62 . . .
is the numerical value of the golden ratio. The set of Fibonacci numbers, 1, 1, 2, 3, 5, 8, 13, 21, . . . can be extended by using the rule that the next number is the sum of the last two previous numbers. Mathematicians have also shown that the ratios of the (n + 1) to the nth term in the set of Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, . . .approach the golden ratio as a limit.
8 Minimal skeletons of radiolarians under the sea
In his celebrated book ON GROWTH AND FORM, D'Arcy Wentworth Thompson cites numerous examples suggesting that the shapes of many living things are largely determined by the application of elementary mathematical principles. Of special interest here is Thompson's study of the manner in which surface tension can determine such shapes. In particular he discusses the tiny marine organisms called radiolarians some of which beautifully illustrate maximizing principles. When radiolarians are alive, they consist of a small mass of protoplasm surrounded by a soap-bubble-like froth of cells. As with soap bubbles and soap films, the fluid in the interfaces of she froth accumulates primarily at the branchings, and the animal acquires an exquisite glassy skeleton by the deposition of a solid out of the fluid. When the animal dies, everything decays except the skeleton. These skeletons provide a striking visualization of the branchings of the froth: the pieces of curves and the vertex points. Some of them also show the shapes of the cell walls.
These six figures portray the microscopic skeletons left after the death and decay of several types of radiolarian, tiny marine organisms that in life consisted of a small mass of protoplasm surrounded by soap-bubble-like froth of cells.
Examples of superlatives using arguments from calculus
Fermat's introduction to calculus solutions
As differential calculus developed it was strongly influenced by individual maximum and minimum problems. Differentiation may be said to have originated in the problem of drawing tangents to curves and in finding maximum and minimum values of functions.
Fermat (1601 1665), "the greatest mathematician of the first part of the 17th century" Although Newton and Leibniz are credited with having invented the calculus, yet Fermat, as early as 1629: this is 57 years earlier than Newton or Leibniz had developed the method that is now standard in the calculus for solving the problem of maximizing or minimizing a function.
Euclid's proof that the square has the greatest area for a rectangle having a fixed perimeter
(For this problem, Fermat used a process based on what we now call DIFFERENTIAL CALCULUS)
To prove: The square (B + C) is greater than an arbitrary rectangle (A + C)
Proof (referring to the figure at the right)
1. 2x + 2u + 2y = Perimeter of the rectangle (A + C) = Perimeter of the square (B + C) = 2x + 2v + 2y
2. So u = v (since 2x + 2u + 2y = 2x + 2v + 2y)
3. So Area A = ux < vy = Area B, (since x < y)
4. So Area (A + C) < Area (B + C)
5. So the Area of an arbitrary rectangle < the Area of a square
Fermat's use of infinitesimals in geometry, and particularly his application of them to questions of maxima and minima,.
Fermat published a threefold process for determining such a maximum or minimum.
1. If f(x) has an ordinary maximum or minimum at x, and if e is very small, then the value of f(x‹e) is almost equal to that of f(x).
2. Therefore, we tentatively set f(x - e) = f(x) and then make the equality correct by letting e assume the value zero.
3. The roots of the resulting equation then give those values of x for which f(x) is a maximum or a minimum.
Consider Fermat's first example
To divide a quantity B into two parts A and B - A such that their product is a maximum.
1. (A-E)[B-(A-E)] now equating this to A(B-A) we get
2. A(B-A) = (A-E)(B-A+E)
3. so 2AE BE E2 = 0 or 2A B E = 0 and setting E = 0 we get 2A =B
This is equivalent to lim [f(x + h) f(x)]/h = o as h->0 which defines the derivative.
Kepler's wine barrels
Kepler's wine barrel problemKepler's "Doliometry," the "barrel calculation." When Kepler, the imperial court astrologer at Linz, married the second time, he bought for the wedding wine from a barrel. To compute the bill, the wine merchant measured the barrel by inserting a foot rule into the taphole S until it reached the lid at D, then he read off the length SD = d and set the price accordingly. This method outraged Kepler, who saw that a narrow, high barrel might have the same linear measure SD as a wide one and would indicate the same wine price, though its volume would be ever so much smaller.
Giving further thought to this method of using d to determine the volume, Kepler approximated the barrel somewhat roughly by a cylinder, with s the radius of the base and h the height. Then he asked: If d is fixed, what value of h gives the largest volume V? V is a polynomial in a. Today we would use a method depending on the derivative to fod that the relation between d and h would have to be 3h2 = 4d2 which is the exact formula Kepler found using a method of indivisibles, that defined a barrel of definite proportions. Kepler noticed that in his Rhenish homeland barrels were narrower and higher than in Austria, where their shape was peculiarly close to that having a maximum volume for a fixed d - so close, indeed, that Kepler could not believe this to be accidental. So he imagined that centuries ago somebody had calculated barrel shapes, as he himself was doing, and had taught the Austrians to construct their barrels in this particular fashion - a very practical one, indeed. Kepler showed that if a barrel did not satisfy the exact mathematical specification 3h2 = 4d2, but deviated somewhat from it, this would have but little effect on the volume, because near its maximum a function changes only slowly. Thus, while the Austrian method of price determination, if applied to Rhenish barrels, would be a clear fraud, it was quite legitimate for Austrian barrels. The Austrian shape had the advantage of permitting such a quick and simple method. So Kepler relaxed and let his suspicions dissipate.
The law of refraction (see figure #6) was published by Descartes in 1637 and proven by Fermat using extrema principles and his Principle of Least Time. Fermat knew that under reflection light takes the path requiring least time and was convinced that nature does indeed act simply and economically. So Heron of Alexandria's use of minima principles for reflection could also be used for refraction.
Fermat's Principle is stated mathematically in several equivalent forms. According to the law of refraction; sina/sinb = va/vb where va is the velocity of light in the first medium and vb in the second. The ratio of' va to vb is denoted by h and is called the index of refraction of the second medium relative to the first, or, if the first is a vacuum, h is called the absolute index of refraction of the nonvacuous medium. If c denotes the velocity of light in a vacuum, then the absolute index h = c/v where v is the velocity of light in the medium. If the medium is variable in character from point to point, then h and v are functions of x, y, and z.
Honeycombs: a study in efficiency
How to cover a honeycomb
Tiling a plane can only be done with tiles in the shape of equilateral triangles, squares, or regular hexagons: there are no other possibilities, since these are the only regular polygons which evenly divides 360°. An example is a cross section of the comb of the honeybee. Around 1712, the Swiss mathematician Samuel Koenig used calculus to prove that the hexagonal shape of the honeycomb's cross section affords optimal utilization of the beeswax. By a famous verdict in the French Academy of Sciences and Letters, it was decided that the bees did not know calculus and when they blindly used it anyhow, "it was due to divine guidance". Bees also know how to cover these cells to maximize the honey capacity
Consider the surface by starting with a regular hexagonal base abedef with side s. Over the base we raise a right prism of a certain height h and with top A BCDEF. The corners B. D, f are cut off by planes through the lines AC, CE, EA, meeting at a point V on the axis VN of the prism, and intersecting Bb, Dd, Ff at X, Y,Z. The three cut-off pieces are the tetrahedrons ABCX, CDEY, EFAZ. We put these pieces on top of the remaining solid such that X, Y and Z coincide with V. Hereby, the lines AC, CE, EA act as "hinges". The faces AXCV, CDEY, EZAY are rhombuses, that is, quadrilaterals with equal sides The new body is the bee's cell and has the same volume as the original prism. The hexagonal base abcde is the open end.
The bees form the faces by using wax. When the volume is given, it is economic to spare wax and, therefore, to choose the angle of inclination, q = angle NVX, in such a way that the surface of the bee's cell is minimized. The problem can be solved using calculus and q is found to equal 54.7o .
Vascular branching of arteries
The blood vascular system consists of arteries, arterioles capillaries, and veins. The transport of blood from the heart through all organs of the body and back to the heart should be as effective as possible. With a minimal energy expenditure, the body should be fed quickly by the constituents of blood. Optimality has to be reached in several ways. For instance. each vessel should be wide enough to avoid turbulence, and erythrocytes should be kept at a size as to minimize viscosity. The reisitance, R, to flow is inversely proportional to the fourth power of the radius. R = k/r4 .
We assume that a main vessel of radius r1, runs along the horizontal line from A to B. A point C should be reached by a branch of a given radius r2 . Using calculus methods we find that the optimal angle is determined by cos q = (r2/r1)4 .
Some geometry, theorems and course syllabi
Six types of Ruled Surfaces
Half Twist Ruled Surfaces
p/q Twist Ruled Surfaces
Saddle (hypar) Surfaces
Geometry of Bridge construction
The Seven Wonders of the Ancient World
The 13 Achimedian semiregular polyhedra
TheoremsThe Mathematician's Quest for Superlatives . . .from geometrical and caculus considerations
The Mathematician's Quest for Superlatives . . .using caculus of variations
Certain Periodic Polar Curves
Monge's Twist-surface Theorems
Hyperpower Function xxx . . .
Theorems of Girolamo Sacceri, S.J. and his hyperbolic geometry
Saccheri's Solution to Euclid's BLEMISH
Course syllabiAnalysis III
Ordinary differential Equations