CHAPTER VII. MAXIMA AND MINIMA. Sect. 1. Explicit Functions of One Variable. SUPPOSE that u is any explicit function of x: the following rule will enable us to determine those values of x which du render u a maximum or minimum. “ Equate to zero or dx infinity: let a be a possible value of x obtained from either du of these equations; then, if changes sign from + to - or dx from to + when, h being an indefinitely small quantity, h and a + h, are substituted successively for x, x = a will correspond respectively to a maximum or minimum value of u: if no such change of sign takes place the value a of x must be rejected. By applying this process to each of the du du roots of the equations 0 and we shall have dx dx determined all the desired values of x." = du Suppose that ¥ (r). («), where y (x) is a function dx of x essentially positive for all possible values of x: then, du instead of we may evidently take v = (r), and treat v dr du just as we should have treated dx The following principle is also frequently useful for the determination of maxima and minima. “Suppose that, for or du du du du any particular value of x, = 0, and that dx dir' dæ?? d v3 are none of them infinite: then, if the first of these differential coefficients which does not vanish, for the particular value of X, be of an even order, u will be a maximum a minimum accordingly as this differential coefficient is du negative or positive." If = 4(x).v, ¥ (0) being an es dx sentially positive function of x, the following modification of this principle in many cases affords considerable simplification. “Suppose that, for any particular value of x, v = 0, and that dv dx' dos da .. are none of them infinite : then, if the first of these differential coefficients which does not vanish, for the particular value of x, be of an odd order, u will be a maximum or a minimum accordingly as this derived function is negative or positive.” ď v In testing by the sign of the first differential co dra efficient of u which does not vanish for a particular value a of x, whether the value of u be a maximum or a minimum, the following consideration will sometimes shorten the process. du and a be a do value of N, one of the factors as w, and its first n - 2 differential coefficients to vanish, the only term of d" u d"-'w, which is to be considered is that involving dar dr-1 causes as all the others vanish when x is put equal to a, so that dra is reduced to one term. The investigation of the maximum and minimum values of u is sometimes facilitated by the following considerations. If u be a maximum or minimum, and a be a positive constant, au is also a maximum or minimum. When u is a maximum or minimum, au"+1 is so also; a but is inversely a minimum or maximum. un+1 If u be a positive maximum or minimum, au’n is also a maximum or minimum. If u be a negative maximum or minimum, au?" will be a minimum or maximum. The same remarks apply to fractional powers of the function u, except that when the denominator of the fraction is even, and the value of u negative, the power of u is impossible. When u is a positive maximum or minimum, log u is a maximum or minimum. This preparation of the function is frequently made when the function u consists of products or quotients of roots and powers, as the differentiation is thus facilitated. Other transformations of u are sometimes useful, but as these depend on particular forms which but rarely occur, they may be left to the ingenuity of the student who desires to simplify the solution of the proposed problem. The roots of this equation are 1, 2, 3, and for x = 1, u is a minimum ; 2, u is a maximum; 3, u is a minimum. for x = for r = The roots O make u neither a maximum nor a minimum; a = a makes u = 0, which is a minimum when n is even, du because changes sign from to + when a – h, a + h, dw are substituted successively for x; and neither a maximum du nor a minimum when n is odd, because is then insusceptible dx of a change of sign. (5) U = x" (a – x)". du = x' (a – X)*-1 {ma - (m + n) x} = 0; dx та the roots of which are x = 0, x = a, and x = m + n x = 0 makes u = 0, a minimum if m be even, and neither a maximum nor a minimum if m be odd. x = a makes u = 0, a minimum if n be even, and neither a maximum nor a minimum if n be odd. d'u d dr da {x*-'(a-x)"-"}{ma-(m+n).x} –(m+n).um-'(a –x)"-', ma and therefore x = makes u a maximum. m + n This is the solution of the problem. To divide the number a into two parts, such that the product of the power of the one by the nth power of the other shall be a maximum. mth = 0. du 1- x? (6) 1 + x da (1 + r*) Since (1 + xo) is essentially positive we have, taking du v instead of da' V = 1 – 22 = (), X = 2, = X = 2, dx U = ?, a minimum. |