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CHAPTER VII.

MAXIMA AND MINIMA.

Sect. 1. Explicit Functions of One Variable.

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SUPPOSE that u is any explicit function of x: the following rule will enable us to determine those values of x which render u a maximum or minimum. “ Equate to zero or infinity: let a be a possible value of x obtained from either of these equations; then, if a changes sign from + to – or from - to + when, h being an indefinitely small quantity, a-h and a + h, are substituted successively for a, 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 roots of the equations =

and I = , we shall have 1. dx

dx determined all the desired values of x.”

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Suppose that = y(x).P(x), where y (x) is a function of x essentially positive for all possible values of x: then, instead of in we may evidently take v = P(x), and treat v

du

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The following principle is also frequently useful for the determination of maxima and minima. “ Suppose that, for

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du du du any particular value of x, qx = 0, and that

do'dur' d3?. 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 or a minimum accordingly as this differential coefficient is

du negative or positive." If = {(x).v, v (v) 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 dv dv ' d e domin... are none of them infinite: then 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.”

d" u In testing by the sign of the first differential coefficient 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.

If .be of the form w, . W, . Wz ... W., and a be a

da value of x, which causes one of the factors as w, and its first n - 2 differential coefficients to vanish, the only term of

“ which is to be considered is that involving , as

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as

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all the others vanish when & is put equal to a, so that is reduced to one term.

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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?n+1 is so also ; but-, is inversely a minimum or maximum.

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If u be a positive maximum or minimum, auan is also a maximum or minimum. If u be a negative maximum or minimum, aun 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.

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The roots of this equation are 1, 2, 3, and

for x = 1, u is a minimum ;
for x = 2, u is a maximum;
for x = 3, u is a minimum.

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: The roots O make u neither a maximum nor a minimum ;

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Q = a makes u = 0, which is a minimum when n is even,

- du . because au changes sign from – to + when a – h, a + h, are substituted successively for w; and neither a maximum

du nor a minimum when n is odd, because is then insusceptible

da of a change of sign. (5) u = x" (a – x)".

= wm-) (a – x )*-1 {ma - (m + n) x} = 0;

du

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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 mth power of the one by the nth power of the other shall be a maximum.

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du 1 - Q (6) U=

da (1 + r) Since (1 + x2) is essentially positive we have, taking v instead of of la

v= 1 – 22 = 0,

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