5.1 Introduction
Now we shall demonstrate how the inequalities that were derived in the preceding chapter can be used to treat an important and fasci nating set of problems. These are
maximization
andminimization
problems.
In the study of algebra and trigonometry, the problems you encoun tered were all of the following general nature: Given certain initial data, and certain operations that were to be performed, you were to determine the outcome. Or, conversely, given the operations that were performed and the outcome, you were to determine what the initial data must have been.
For example, you met three workmen, eager and industrious
A,
plodding and conscientious B, and downright lazy C. They were put to work digging ditches, constructing swimming pools, or building houses, and the problem was always that of determining how long it would take the three of them together to perform the job, given their individual efficiency ratings-or, given the time required for all three together, and the rates of effort of two of the three workers, it was required to find the rate of the third.
Sometimes you were given triangles in which two sides and an included angle were known, and sometimes you were given three sides
80 A N I N T R O D U C T I O N T O I N E Q U A L I T I E S
of the triangle. In each case, the problem was that of determining the remaining sides and angles.
The foregoing kinds of problems are simple versions of what may be called
descriptive problems.
Here, we wish to consider quite different types of problems. We shall consider situations in which there are many ways of proceeding, and for which the problem is that of determining
optimal choices.
Questions of this nature occur in all branches of science and consti tute one of the most important applications of mathematical analysis. Furthermore, much of physics and engineering is dominated by prin ciples asserting that physical processes occur in nature as if some quantity, such as time or energy, were being minimized or maximized. Many processes of this sort can be treated in routine fashion by that magic technique of Fermat, Leibniz, and Newton, the calculus. Supposedly, in the 1 7th century, men who knew calculus were pointed out in the street as possessing this extraordinary knowledge. Today it is a subject that can be taught in high school and college, and requires no exceptional aptitude.
When you take a course in calculus, you will find that many of the problems treated here by algebraic processes can also be solved quite readily by means of calculus. It will then be amusing to solve them mentally by means of the techniques presented here, but to follow the formalism of calculus in order to check your answer.
Each technique, calculus as well as the theory of inequalities, has its own advantages and disadvantages as far as applications to maxi mization and minimization problems are concerned. It is rather characteristic of mathematics that there should be a great overlapping of techniques in the solution of any particular problem. Usually, a problem that can be solved in one way can also be solved in a num ber of other ways.
5.2 The Problem of Dido
According to legend, the city of Carthage was founded by Dido, a princess from the land of Tyre. Seeking land for this new settlement, she obtained a grudging concession from the local natives to occupy as much land as could be encompassed by a cowhide. Realizing that, taken literally, this would result in a certain amount of overcrowding, she very ingeniously cut the skin into thin strips and then strung them out so as to surround a much larger territory than the one-cowhide area that the natives had intended.
M A X I M I Z A T I O N A N D M I N I M I Z A T I O N 8 1 The mathematical problem that she faced was that of determining the closed curve of fixed perimeter that would surround the greatest area; see Fig. 5 . 1 .
I n its general form, the problem is much too complicated for u s to handle. As a matter of fact, at our present level we don't even know how to formulate it in precise terms, since we have no way of express ing what we mean either by the perimeter of an arbitrary curve or by the area enclosed by this curve. Calculus suggests definitions and provides analytic expressions for these quantities, and a still more advanced part of mathematics called the
calculus of variations
pro vides a solution to the problem.As one might guess, the optimal curve in Dido's problem is a circle: Let us here, however, concentrate on certain simple versions of the problem that we can' easily handle by means of the fundamen tal results we have already obtained in the theory of inequalities. If you would like to pursue the study still further, you should read the interesting tract
Geometric Inequalities
by N. D. Kazarinoff, also in this series.5.3 A Simplified Version of Dido's Problem
Let us suppose, for various practical considerations, that Dido was constrained to select a rectangular plot ofland, as indicated in Fig. 5.2.
Figure 5.1. The problem of Dido
X
X
Figure 5.2.
Simplified version of the problem of Dido With the lengths of the sides of the rectangle designated by x and y, respectively, we see that the length of its perimeter is given by the
algebraic expression
(5. 1)
L
= 2x+
2y ,and its area is represented by the formula
82 A N I N T R 0 D U C T I 0 N T 0 I N E Q U A L l T I E S
Since x andy represent lengths, they necessarily are nonnegative quantities. The fact that the perimeter 2x + 2y is equal to L implies that x andy must satisfy the inequalities
L L
(5.3)
2
�X
�0 '
T
� y �
0 .
It is clear, then, that the area A = xy cannot be arbitrarily large. Indeed, from the inequalities (5.3) and the expression (5.2) for the area, we see by Theorem 2. 5 of Chapter 2 that A cannot exceed the value L214 ; that is, we have
(5.4) 4 � xy= A . L2
How are we going to determine the dimensions that yield the maximum area?
Referring to the inequality connecting the arithmetic mean and geometric mean of two quantities, we observe that
(5.5)
for all nonnegative numbers x andy. Since, in this case, we have x + y = L/ 2 , the inequality (5.5) yields the relationship
(5.6) or L2
16 � xy = A .
Consequently, we see that our first rough bound for the area, L214 , obtained in (5.4), can be considerably "tightened." We can, however, go much further. Recall that we have established that the equality in (5.6) holds if and only if x = y . In our case, this means that the new upper bound, L2 I 16 , is
attained
if and only if x = y . For all other nonnegative choices of x andy satisfying the relationship (5. 1 ), the area is less than L2 I 1 6 .What does the foregoing analysis tell us? It tells us that in no case can the area of a rectangle with perimeter L exceed the quantity L2 I 16 , and that this maximum value is actually reached if and only if we choose the sides equal to each other, and therefore equal to the quantity L14 .
Thus, by purely algebraic means we have achieved a proof of the well-known and rather intuitively obvious fact that