Page:EB1911 - Volume 17.djvu/938

919
MAXIMA AND MINIMA

the action of gravity, proposed in 1696, gave rise to a new kind of maximum and minimum problem in which we have to find a curve and not points on a given curve. From these problems arose the “ Calculus of Variations.” (See VARIATIONS, CALCULUS or.)

The only general methods of attacking problems on maxima and minima are those of the differential calculus or, in geometrical problems, what is practically Fermat's method. Some problems may be solved by algebra; thus if y=f (x)+ ¢> (x), where f (x) and qb (x) are polynomials in x, the limits to the values of y may be found from the consideration that the equation y4> (x) —f (x)=o must have real roots. This is a useful method in the case in which ¢> (x) and f (x) are quadratics, but scarcely ever in any other case. The problem of finding the maximum product of n positive quantities whose sum is given may also be found, algebraically, thus. If a and b are any two real unequal quantities whatever {%(a~|-b)}'>ab, so that we can increase the product leaving the sum unaltered by replacing any two terms by half their sum, and so long as any two of the quantities are unequal we can increase the product. Now, the quantities being all positive, the product cannot be increased without limit and must somewhere attain a maximum, and no other form of the product than that in which they are all equal can be the maximum, so that the product is a maximum when they are all equal. Its minimum value is obviously zero. If the restriction that all the quantities shall be positive is removed, the product can be made equal to any quantity, positive or negative. So other theorems of algebra, which are stated as theorems on inequalities, may be regarded as algebraic solutions of problems on maxima and minima.

For purely geometrical questions the only general method available is practically that employed by Fermat. If a quantity depends on the position of some point P on a curve, and if its value is equal at two neighbouring points P and P', then at some position between P and P' it attains a maximum or minimum, and this position may be found by making P and P' approach each other indefinitely. Take for instance the problem of Regiomontanus “ to find a point on a given straight line which subtends a maximum angle at two given points A and B.” Let P and P be two near points on the given straight line such that the angles APB and AP'B are equal. Then ABPP' lie on a. circle. By making P and P' approach each other we see that for a maximum or minimum value of the angle APB, P is a point in which a circle drawn through AB touches the given straight line. Theretare two such points, and unless the given straight line is at right angles to AB the two angles obtained are not the same. It is easily seen that both angles are maxima, one for points on the given straight line on one side of its intersection with AB, the other for points on the other side. For further examples of this method together with most other geometrical problems on maxima and minima of any interest or importance the reader may consult such a book as ]. W. Russell's A Sequel to Elementary Geometry (Oxford, 1907).

The method of the differential calculus is theoretically very simple. Let u be a function of several variables supposed for the present independent; if u is minimum for the set of values xl, xz, xg, . . .x, ., u+6u, when xl, xg, x3, ... x, . receive 'small 6x¢, . ..¢5x, .; then 6u must have the same sign values of 5x1, éxg, . 6x, ..

MW au=z§ f;ax, +; z%f, ax,2+2z%ax, ax, . + . xi, 902, 1263, ~ - -rm

a maximum or

and u becomes

increments 6x1,

for all possible

The sign of this expression in general is that of E(5u/8x1)6x1, which cannot be one-signed when xl, xg, . . . x, , can take all possible values, for a set of increments 5x1, 6x2, . . . 6x, , will give an opposite sign to the set-Bxl, -6x2, . . -éxf.. Hence 2(5u/5x1)6x1 must vanish for all sets of increments Bxl, . . Bxn, and since these are independent, we must have 514/6x, =o, 614/5x2=o, 5u/6x, .=o. A value of u given by a set of solutions of these equations is called a “ critical value ” of u. The value of 6u now becomes 2 2

é 2% 6x, '+2 21%-c25x,6x2+ . .;

for u to be a maximum or minimum this must have always the same sign. For the case of a single variable x, corresponding to a value of x given by the equation du/dx=0, u is a maximum or minimum as dfu/dx” is negative or positive. If dzu/dxf vanishes, then there is no maximum or minimun unless dau/dx' vanishes, and there is a maximum or minimum according as d4u/dx* is negative or positive. Generally, if the first differential coefficient which does not vanish is even, there is a maximum or minimum according as this is negative or positive. If it is odd, there is no maximum or minimum. In the case of several variables, the quadratic 6214 5214

E- 2 2 T .

536125361 'l' Eaxlaxz 5.'JC15xg +

niust be one-signed. The condition for this is that the series of discriminants

an, an (liz |, Ufu lliz Gia . . .

021 1122 a2l 022 a23

G31 G32 as a

where ap., denotes 5'u/5a, ,6aq should be all positive, if the quadratic is always positive, and alternately negative and positive, if the quadratic is always negative. If the first condition is satisfied the critical value is a minimum, if the second it is a maximum. For the case of two variables the conditions are &. & (W 2

53012 63C22> 5x| 5x2>

for a maximum or minimum at all and 5%/6x12 and 5214/6x22 both negative for a maximum, and both positive for a minimum. It is important to notice that by the quadratic being one-signed is meant that it cannot be made to vanish except when éxl, éxg, . . . 6x, . all vanish. If, in the case of two variables, Ee. '22, (52 ~)”

6x12 63622 5361513

then the quadratic is one-signed unless it vanishes, but the value of u is not necessarily a maximum or minimum, and the terms of the third and possibly fourth order must be taken account of. Take for instance the function u x3—xy*-i-ya. Here the values x=o, y=o satisfy the equations 6u/6x=o, 6u/5y=o, so that zero is a critical value of u, but it is neither a maximum nor a minimum although the terms of the second order are (6x)2, and are never negative. Here 6u=6x”-5x6y2+5y3, and by putting 5x=o or an infinitesimal of the same order as 5312, we can make the sign of éu depend on that of 6313, and so be positive or negative as we please. On the other hand, if we take the function usxz-xy2+y', x=o, y =o make zero a critical value of u, and here 6u=5x2-6x5y2+6y', which is always positive, because we can write it as the sum of two squares, viz. (Ex-=};6y2)2-I-i-6y4; so that in this case zero is a minimum value o u.

A critical value usually gives a maximum or minimum in the case of a function of one variable, and often in the case of several independent variables, but all maxima and minima, particularly absolutely greatest and least values, are not necessarily critical values. If, for example, x is restricted to lie between the values a and b and ¢>'(x)=o has no roots in this interval, it follows that ¢'(x) is one-signed as x increases from a to b, so that 4>(x) is increasing or diminishing all the time, and the greatest and least values of ¢(x) are ¢(a) and ¢(b), though neither of them is a critical value. Consider the following example: A person in a boat a miles' from the nearest point of the beach wishes to reach as quickly as possible a point b miles from that point along the shore. The ratio of his rate of walking to his rate of rowing is cosec a. Where should he land?

Here let AB be the direction of the beach, A the nearest oint to the boat O, and B the point he wishes to reach. Clearly he must land, if at all, between A and B. Suppose he lands at P. Let the angle AOP be 0, so that OP=a sec 0, and PB =b-a tan 6. If his rate of rowing is V miles an hour his time will be a sec 0/V+ (b-a tan 0) sin a/V hours. Call this T. Then to the first power of 50, 6T=(a/V) sec20 (sin 0-sin a)60, so that if AOB>a, BT and 66 have opposite signs from 0=o to 0=u., and the same signs from 9 =a to 0=AOB. So that when AOB is > a, T decreases from 0=o to 6 =a, and then increases, so that he should land at a point distant a tan o. from A, unless a tan a>b. When this is the case, 6T and 50 have opposite signs throughout the whole range of 0, so that T decreases as 0 increases, and he should row direct to B. In the first case the minimum value of T is also a critical value; in the second case it is not.

The greatest and least values of the bending moments of loaded rods are often at the extremities of the divisions of the rods and not at points given by critical values.

In the case of a function of several variables, xl, x2, ...x, ., not independent but connected by m functional relations u1=o, mio, . . ., u, ,, =o, we might proceed to eliminate m of the variables; but Lagrange's “ Method of undetermined Multipliersf is more elegant and generally more useful.

We have |5u1=o, 6u2=0, . . ., ¢\$u, ,, =o. Consider instead of éu, what is the same thing, viz., 5u+)u1+>u2 + +), ,,6u, ,., where M, M, . . 7, ,, , are arbitrary multipliers. The terms of the first order in this expression are 