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In the former of which cases r"(x) is negative, and the corresponding value of t(x) is a maximum; and in the latter r"(x) is positive, and the corresponding value of F(ct) is a minimum.

150.] This method however of determining such singular values of functions is not applicable whenever the value of x which makes r\x) = 0, also makes v"(x) = 0; in which case, as well as in all others, the following method may be employed.

Let Y(x) be the function of which the maximum and minimum values are to be determined; then, by (21), Art. 116,

and if h is infinitesimal, ¥'(x + 0h) becomes v\x).

Suppose now that x0 is a value of x, such that F(#o) is a maximum or a minimum; then if we consider the values of ?(x) which are immediately adjacent to r(xu) on both sides of it, we shall be able to detect a criterion of such singular values. For this purpose let us consider r(X(, + h) and F(<r0—A) when h is infinitesimal: then if F (x0) is a maximum, F + h) F (x„) and r(x0 h)— v(x0) are both negative: and if F(a?0) is a minimum, these differences are both positive. Now if does

and thus these differences have different signs. But this is inconsistent with F(#0) being a maximum or minimum ; and therefore we conclude that F(x0) is not a maximum or a minimum if F' (x0) does not vanish.

Suppose however that F'(x0) = 0, and that F"(.ro) does not vanish; then, by equation (22), Art. 116,

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F (•*■<,+ h) = J-tjf'vo);

and this does not change sign with h; if therefore v"(x0) is positive, r(x0) is less than both ¥(x0 + h) and F(#0A)» that is, F(x0) is a minimum; and if r"(xu) is negative, r(x0) is greater than both r(x» + h) and F(^0 k), that is, v(x0) is a maximum; whence we conclude,

If r'(x0) = Oj and *"(xo) docs not vanish, F(-r0) is a singular value; and if F"(;to) is negative, F(.r0) is a maximum; and if K"(,t'o) is positive, v(x0) is a minimum.

Again, if F"(.r0) = 0, and F"'(x0) does not vanish, and if h is infinitesimal,


F (x0 + h)-v Oo) = F'"(,ro)'

in which case, as h3 changes its sign with h, it is plain that there is no maximum or minimum value; but if t"'(,x0) = 0, and i''v(.r„) does not vanish, then


F (x0 + h) - v (xtt) = 1 Tf(x0);

in which case, as before, v(x0) will be a maximum or minimum value of F(#), according as F/V(a0) is negative or positive.

And thus generally if the value #0, which makes = 0, so affects F"(#), r"'(x), ... up to v"-1(x), that all vanish, but that V(x0) docs not vanish; then we have


r(x0 + h) - Y(x0) = Y2~3—« 1

and if n is an odd number, there is no maximum or minimum value; but if n is an even number, F(#0) is a maximum if F"(xo) is negative, and a minimum if F"o0) is positive.

In the application of this theory to questions of geometrical maxima and minima, it will subsequently appear that figs. 12 and 13 correspond to the analytical conditions of every derivedfunction vanishing, when x = x0, up to one of an odd order inclusively, and of the next derived-function of an even order remaining finite; and figs. 14 and 15 correspond to the condition that the derived-function, which is the first not to vanish, is of an odd order.

Ex.1. v(x) = e* + 2 cos x + e~x,

T'(x) = ex 2 Bin x e~x = 0, if x = 0,
r"(x) = ex2cosa? + c-x = 0, - - - ,
r"'(j?) = ex + 2 sin x e— = 0, - - - ,
r^ix) = ex+ 2 cos a? + e~* = 4, - - -;

therefore if x = 0, v(x) = 4, which is a minimum value, because the fourth derived-function, which is the first not to vanish, is positive.

151.] We subjoin some examples of problems on maxima and minima, the study of which will be sufficient to enable the reader to apply the methods to other similar ones.

Ex. 1. To divide a given number a into two parts such that the product of the nth power of one, and the with power of the other, may be a maximum.

Let a be the given number, of which let x be one part, and therefore a—x is the other; and we have

v(x) = x" (a—x)m;

.•. /(a?) = x"-1 (a—x)TM-1 {na nx mx),

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= 0, if x = 0, and, if n is an even number, changes sign from

— to +, which indicates a minimum; but if n is odd, r\x) does not change sign, and there is no corresponding maximum or minimum;

= 0, if x = a, and, if m is an even number, changes sign from

— to +, which indicates a minimum; but if m is odd, p'(o?) does not change sign, and there is no corresponding maximum or minimum;


= 0, if x = , and changes sign from + to —, which in

m + n ° °

(a )
tn -(- n'

Ex. 2. To find the number the ratio of which to its logarithm is a minimum.

Let x be the number; then

F (x) = = ;which is to be a minimum;

log x

,, , loea? —1
(log xY

= 0, when x = e, and changes sign from — to +, which indicates a minimum; viz. r(x) = c.

Ex. 3. To find the number of equal parts into which a given number a is to be divided, so that their continued product may be a maximum.

Let x be the number of parts; and thus each part = -; and therefore, if r(x) = the product of them,

,<•) = (J)*;
.•. logr(.r) = x {log a — log x};

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therefore p'(,r) = 0, if x = ", and changes sign from + to —; which indicates a maximum; therefore each part = e, and the


product of all = (e)'.

Ex. 4. To inscribe in a given circle the greatest isosceles triangle.

In fig. 16, let the vertex of the triangle be at the extremity A of the diameter of the circle; let the radius of the circle = a; and let p and p' be the other angular points of the triangle; therefore Mp = Mp'; let Cm = I, Mp = y; then the equation to the circle is . . . .

the area of the triangle = F(.t) = Am x Mp,

= (a+x) (a2—*2)*; a2ax— 2x2 (a + *)*(a — 2x)

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= 0, if x = ^, and the sign changes from + to —, which in

dicates a maximum, of which the value is — a2.


r'(x) also vanishes when x = a, and when x = a; but as the sign of it in both cases passes from + v to +, and as the geometrical meaning of the values is plain enough, we must reserve the consideration of such critical values to a future part of the work.

Ex. 5. To inscribe the greatest rectangle in a semi-ellipse.

In fig. 17 let A'p'bpa be the given semi-ellipse, and Mpp'm' the rectangle inscribed in it, which is to be a maximum; let c M = x, M p = y; and let the equation to the ellipse be

.-. the rectangle = p(.r) = M'mxmp,

= 2 c M x M p,

= 2-x(a*-x2)t;

_A a*-2xi
.-. v'(.c) = 2 -

a (a2J?2)4

= 0, if x = H—^5, and changes sign from + to —, which

indicates a maximum, and r(x) becomes ab; and therefore the greatest rectangle is one-half of that contained by Aa and Cb.

Ex. 6. The whole surface of a cylinder being given, it is required to find its form when the content is a maximum.

Let x the radius of the base, y = the height of the cylinder; .•. nx*y = the content, and 2irx2 + 2irxy = the whole surface. Let the given surface be 2 wo2, and let the content be v(x);

.-. 2irxt+2irxy = 2ira2, and y =


r(x) = nx2y, = Ttx (a2x2);
.-. r'(x) = ir(a2-3.r2),


= 0, if x = —and changes sign from + to —; which indi


cates a maximum; in which case y = the height = —^, 2na3

and the content = —- -.

Ex. 7. Given the content of a cone; it is required to find its dimensions when the whole surface is a minimum.

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