SECTION CXIII.

On Maxima and Minima.

Any algebraic expression containing a letter, is called a function of that letter: thus, a x — - x2, ax, bcx + x2, &c., are functions of x.

It is sometimes necessary to know what value of a letter will render a given function the greatest or least possible; that is, greater or less than any other value of the letter will make it.

1. It is required to divide the number 10 into two such parts, that their product will be greater than the product of any other two parts of 10.

It is evident that there are some two parts of 10 which will give a greater product than any other two parts will give.

1st. Let

Then
And

2nd. Assume

x= one part of 10,

10-x the other part,
10x-x2= the product of the parts.

x+h one part of 10,

Then 10-(x+h) the other part,

And 10(x+h)-(x+h)

=

=

=

the product of the two parts.

one part of 10,

-h) the other part,

Then 10 (x
And 10 (x-h)-(x — h)2 =

the product of the two parts. Here observe h may represent any quantity or number whatever, however small or large. It is altogether indefinite. Now 10(xh)-(x + h)2 = 10x-x2+(10-2x) h—h2 And 10x-h—(xh)2 = 10x-x2-(10-2x) h—h2

When, therefore, we assume x = one part of 10, 10 x - x2 will represent the product. Assuming x+h for one part, 10x-x2+(10-2x) h-h2 will represent the product. Assuming xh for one part, 10x-x-(10-2x) h—h2 will represent the product.

The products of the parts, then, take the following forms, according as we take x for one part, or a number greater than by h, or less than x by h: viz.

Now we wish to find such a value of x as will make the expression (1) greater than either (2) or (3.) This value will be one of the parts of 10 required; for, we cannot increase it or diminish it, without making the product of the two parts less.

Make the term multiplied by h, in the expressions (2) and (3) = 0, and then the three products become

And, it is evident that the (5) and (6) are now each less than the first, by h. Hence, making (10-2x) h=0 will give the We have, then, two factors, 10-2 x and h, the product of which, is equal to nothing. One of these factors must be = 0; but, h is some real number.

result required.

Hence

10-2x=0
2x= : 10

x= 5= one part.
the other part.

10- -x= 5

10x-x2 = 25

=

Twenty-five, then, is the greatest product which can be made by any two parts of 10; for, it will be seen by inspecting (4,) (5,) (6,) that if we vary the part 5 by any number h, the product will be diminished by the second power of h. Reviewing the process we find that we assumed x for one part of 10, and found the product represented by

10x-x2

Then substituting x+h and x-h for x, developing and collecting together the terms multiplied by the same powers of h, we have

And

10x-x2+(10-2x) h—h2
10x-x(10-2x) h-h2

Making the terms which are multiplied by the h=0, we find the value of a required.

If we have any function of a letter, we employ the same process, in order to make the function the greatest possible. To show that this process is of general application, we use the following propositions:

PROPOSITION 1st.-If we diminish any fraction of a unit, its second power will be diminished in a greater ratio than the fraction itself; and the smaller the fraction, the greater will be

the difference, in the ratio of diminution, of its first and second powers.

to

Illustration.-Take; its second power is. If we diminish it, we shall have, and its second power Here we see, that by diminishing the fraction, its second power was diminished to what it was before. Take; its second power is. If we reduce the its second power is reduced to T. Here, by reducing the fraction to of its value, its second power is reduced to of its value. It is evident, that any higher power of the fraction will be diminished in a still greater ratio.

PROPOSITION 2nd.-If we have any two numbers, a and c, we can assume h so small, that a h will be greater than ch2. Illustration.-Let a = 10, and c 100.

we could diminish this assumed value; and, as its second power is diminished in a much greater ratio, we shall easily find a value for h that will make

ahcha

PROPOSITION 3rd.—It is evident, that on the same principle, we can assume some value for h, which will make a h2> than ch3; and indeed that we can make any number multiplied by h, greater than the sum of any limited numbers multiplied by higher powers of h; and, moreover, we can assume some value of h that will make any limited number or quantity multiplied by any power of h, greater than limited sum multiplied by higher powers of h.

any

Let us resume the expressions (1,) (2,) (3,) in order to present a view of them which may be applied to subsequent questions; we shall substitute a for 10.

Subtract (1) from (2) and (3) and we have the remainders

We must make these remainders both negative; for they have each resulted from subtracting a greater from a less; and they must be negative, whatever h may be.

Now, if a-2x were equal to any real positive number, we might assume h so small, that that number multiplied by h would be greater than h2. [PROP. 2.] This would make (A) positive; and consequently, make (1) less than (2.) a-2x cannot be any negative number; for, since it is subtracted in (B) it will become plus; and, consequently, (1) less than (3.) Since, then, a-2x can be no real number, positive or negative, it must be

= 0.

=

α-2x=0

a=2x

x=

α

When we make a 2x=0, we see, that the expression (1) is greater than (2) and (3,) by h2; for, the (2) and (3) is the same as the (1); but if the h were plus, then (1) would be less than either (2) or (3) so that the value of x found, would give the least value of the function instead of the greatest. Hence, after we put the terms multiplied by h, = = 0, we must examine whether the remaining terms are — If they are - the value of the letter found will give the greatest value of the function; if +, the least.

or +.

2. Divide the number a into two such parts, that one multiplied by the second power of the other may be the greatest possible.

x= the other.

а х2 - 23

the product of one by the square of the other. Assuming xh for one part, we should have the same product = a(x+ h)2 — (x + h)3.

Assuming xh for one part, the product would be a (x — h)3 — (x — h)3

Developing the two latter expressions, and collecting the terms multiplied by the same powers of h, we have

a x2

a x2x2+(2 ax-3x2) h + (a-3 x) h2-h3
ax2x2-(2 ax-3x2+(a—3x) h2 +2

Subtracting (1) from (2) and from (3), we have

(A) (2 ax-3x2) h + (a—3 x) h2 — h3 And (B) (2 a x − 3 x2) h + (a — 3 x) h2 + h3

These expressions, (A) and (B) must be both negative, whatever h may be; but it is not possible to make (A) negative, if the terms multiplied by h make any positive number; for, since it is multiplied by the first power of h, the h may be assumed so small, that the term [PROP. 2.] (2 ax-3x2) h will be more than all the other terms; so that there would be a positive whole number remaining, if both the other terms were subtracted.

The term multiplied by h cannot be any negative number, since it would become positive in (B,) and be more than all the other terms.

The expression (1) will be the greatest or least, according

as (C) and (D) are negative or positive. Substitute

x in (C) and (D,) and they become

which are evidently both negative.

2 a

for 3

By the same process of reasoning, it may be shown, in any case, that the term multiplied by the first power of h must be made = = 0, in order to obtain the greatest or least value of the function. Whether this value of the letter gives the greatest or least value of the function, may be determined by examin