the term ax will be greater than the sum of all the consecutive terms.

Hence it will also be easy to determine the value of x in such manner that any term of a proposed series becomes greater than the sum of all the consecutive terms. If, for example, it were proposed to make the terms pa" greater than the sum of all those which follow it, we need only, according to the preceding make x = _p s designating the greatest coefficient of all those which folp+s

low p.

37. In a function of x, represented by y, instead of x, putting successively x+h and x - h, and designating the corresponding values of y by y, and y,

Now instead of x, putting in these two expressions, a value represented by a, for which p becomes =0, and designating the values of y,, y, p, q, r changed by the supposition of a = a, by

If h is made so small that q'

h2

+ p 2

2
h2

1 2

h3

1 2

h3

1 2

(see art. 16), we obtain

is greater than the sum of all the consecu

tive terms (36), and if we suppose q' positive, the expression

y', as well as y' will therefore be greater than y', that is to say y' yxa will be less than the value which immediately precedes and follows it, or (35) y is a minimum when x = a. If, on the contrary, on the same supposition for the value of h, we make q' negative, y,, as well as y' will be less than y', or (35) y is

is

a maximum when x = a. The condition under which y becomes a minimum, is therefore that instead of x, we suppose a value = a which makes p vanish and renders q positive; and the condition under which y becomes a maximum, is that instead of x, we suppose a value = a which makes p vanish and renders q negative. Therefore, in order in a proposed function, to examine whether y is a maximum or a minimum, and for what value of x it obtains, we must make p = 0, and determine a by this equation. If q becomes positive for the value found of x, which we designate by a, y is a minimum for x = a; if, on the contrary, q becomes negative by this supposition, y is a maximum for x = a.

Ꮖ

38. It is manifest that we must examine for each real value of a which we obtain from the equation p O, the sign by which q is affected, and that only the imaginary values of a may be neglected in the equation p = 0; it also hence results that a function may have several maxima and minima.

If q does not contain x, which will be the case when the equation p = = 0 includes a only in the first power, q indicates immediately by its sign, whether the function is a maximum or a minimum.

A few examples will elucidate what has been said.

Therefore the proposed function is for x = +a a minimum, and for x = — За a maximum. The value of the minimum (as may easily be found by the substi

tution of the value of x in the proposed function) is =

maximum is = + 9a3.

5

a3, and that of the

two last values being imaginary, we need only consider the first, by which q becomes = 6a2. The proposed function is therefore a minimum for x =

(3.) Let the function proposed be that already considered in art. 35.

The function is a minimum for the value of x=m, which we deduce from p = 0, which results immediately from the positive sign by which q is affected.

39. If q as well as p, vanishes for x = a, the two series of No. 37 change into the following:

greater than the sum of all the consecutive terms (36), it will, because of the opposite signs by which it is affected in the two series, make y', greater and y' less than y'. The function consequently will be neither maximum nor minimum in the case of p and q vanishing for x = a. But if, on the contrary, on the supposition of a = a, r as well as p and q becomes equal to zero, a circumstance in which we have

y' will be a maximum or a minimum, according as s is negative or positive for dmy x= a. Designating generally by the first of the differential coefficients of dxm

the function y which, under the supposition of x = a, does not vanish; the function y will be for x = a, neither a maximum nor a minimum when m is an odd number; but it will be a maximum or a minimum when m is an even and a

number; that is to say, a maximum, if

dmy dam

dmy dam

minimum if becomes positive for x = a.

Let us, for example, consider the function

becomes negative for x = a,

From p = 0, we obtain x = a; for this value of x all the successive differential coefficients vanish, as far as that of the order n the proposed function may therefore, according to what has just been said, be a maximum or a minimum, if n is an even number: but it is in fact only a minimum, because the first of the differential coefficients which do not vanish, is = n (n − 1) positive.

40. We now present some problems on this subject.

1, therefore

Ex. 1. A given number m is to be divided into two parts, such that their product shall be a maximum. In this, as in the following problems, we must in the first place determine the independent x and the function y. Let x be one of the two parts, and y the product of the two parts susceptible of a maximum, and which depend on x. Now, if one of the two parts is = x, the other is =m— x ; therefore, according to the problem,

m

x).
q

the value of x = found from the equation p = 0, gives consequently a

maximum, because of the negative sign of q (art. 38,)

The solution of the proposed problem consists therefore in dividing the given number m into two equal parts.

It is at once obvious that this problem includes also the following: "Let it be proposed to divide a given right line into two parts, in such manner that we may with it describe a rectangle whose area shall be a maximum." The solution gives a square whose area will in fact be the greatest among all rectangles of the same perimeter.

Ex. 2. Within an angle BAC, whose sine is = a, is given a point 0; it is proposed to draw through O a right line to the two lines bounding the angle, in such manner that the triangle so formed shall be a minimum.

The position of O being given, we may consider the lines OB, OC, drawn through the point O, parallel to the two sides of the angle, as given; denote them by a and b. Sup

B

D

A CE

pose the line DE drawn through O to answer the required condition, namely, to be, among all the lines drawn through O, that which includes the smallest triangle DAE.

Put DB = x, then x ax + b: AE, or AE =

a (b + x);

ac

Now, the triangle being the function which is susceptible of a minimum, let it be denoted by y.

We have thus

The first value x =

brenders q positive, and consequently indicates a

minimum whose value is = 2aab, that is to say double the parallelogram ABOC. For the construction of this said triangle, make BD = AB = b, and then from the point D, which will be thus determined, draw a right line through O. The other value x = - b renders q negative, and thus indicates a maximum. But, it is manifest, that the triangle cannot be a maximum; for if a diminishes, the triangle augments more and more, and increases continually, as may be plainly seen from the figure (compare art. 35): therefore x = - - b cannot indicate the greatest triangle. It is nevertheless certain (according to art. 37) that aa (b + x)2 the value of the function y = answers to a maximum if considered in itself, or algebraically, that is, without reference to the proposed problem. To make this still plainer, it will be well to construct the function geometrically, as is often done, that is to say in a curve (in which x and y are considered as coordinates). We thus obtain a hyperbola in which the points x = +b evidently indicate the least positive ordinate and the least negative ordinate, that is to say, the ordinates of which one is a maximum, the other a minimum.

2x

It will be evident that x = - b gives in fact a maximum of the function regarded in itself, but not a solution for the proposed problem. We have therefore the same case here as in the problem, which lead to certain equations, in which some values answer to the equations, but not to the solution of the proposed problem.

The particular determination which the solution takes may be easily found if the given angle is a right angle, where a = 1; and also when the point O is so taken in the midst of the angular space that a = b.

Ex. 3. It is proposed to find in the line PQ the point O, such that the sum of its distances from the two given points M and N, taken out of the line shall be a minimum.

Let the perpendiculars MP and NQ be drawn from M and N, and let them be denoted by a and b; let PQ = c; let also OP: =x, we then

have

M

P

y = MO + NO = (a2 + x2)3 + [b2 + (c − ∞)2]3

From p = 0, we obtain two values for x, namely, x =

-ac

ac

and x = b + a'

But it will be at once

both of which, making q positive, indicate minima. obvious, from the geometrical construction, that there will be only one minimum. To this end let NQ be prolonged until QN' = QN, and draw MN'. The point of intersection O will be the point sought, since the sum

MO + N'O MO + NO

is manifestly less than the sum of any other two lines, for example MO' + N'O' = MO' + NO'. Finding the value of PO, by means of the similar triangles MPO and N'OQ, we obtain for it the first value of x, found above. That first value of x passes evidently to the other if a is taken in a negative sense, that is, so that the two points M and N are found on opposite sides of the line PQ.

ac

b a'

What will be the value of the minimum? and what particular determinations does the solution undergo when a = = b?

Ex. 4. Out of all the right cones, whose surface = m2, to find that whose volume is a maximum.

Ex. 5. Required to find, among all right cylinders whose whole surface =m2, that whose volume is a maximum: take also the case when the curve surface + one end = m2.

Ex. 11. Find an arc such that the rectangle under its tangent and the cosine of its double may be a maximum or a minimum.

Ex. 12. In a spherical triangle, right angled at B, given the angle A, to find the sides so that their difference b c shall be a maximum.

Ex. 13. Find the least parabola which shall circumscribe a given circle. Ex. 14. Given the base of a plane triangle, and its altitude greater than the base, to determine it so that the vertical angle shall be a maximum.

Ex. 15. Having given a line, AB, in length, and another, CPD, in position, to find the point P at which the line AB will appear under the greatest angle possible.

Ex. 16. To find on the line which joins two lights of given intensities, the point which will be least illuminated by both together *.

*For other problems on maxima and minima, see pages 237-239 of this volume.