3a(a2-b2)>2d(c2 —d2).

The same principle is true for division.

But when the two members of an inequality are multiplied or divided by a negative quantity, the inequality subsists in a contrary sense.

Take, for example, 8>7; multiplying by -3, we have

Therefore, when the two members of an inequality are multiplied or divided by a number expressed algebraically, it is necessary to ascertain whether the multiplier or divisor is negative; for, in this case, the inequality would exist in a contrary

sense.

In the problem of No. 104, from the inequality

a2q(a2-b2)>s2 (a2 —b2),

We have been able to deduce q>

$2

by dividing by

a2 (a2-b2), because we have supposed a>b, or a2—b2, positive.

It is not permitted to change the signs of the two members of an inequality unless we establish the resulting inequality in a contrary sense; for the transformation is evidently the same as multiplying the two members by -1.

Transformation by squaring. Both members of an inequality between absolute numbers can be squared, and the inequality will exist in the same sense.

Thus from 5>3, we deduce 25 >9; from a+b>c, we find (a+b)2 >c2.

But when both members of the inequality are not positive, we cannot tell before the operation is performed, in what sense the resulting inequality will exist.

For example, -2<3 gives (—2) or 4<9; but —3—5 gives, on the contrary, (-3) or 9<(-5)2 or 25.

We must then, before squaring, ascertain whether the two ombers can be considered as absolute numbers.

Transformation by extracting the square root.

The square

root of the two members of an inequality between absolute numbers may be extracted, and the inequality will exist in the same sense between the numerical values of the square roots.

We will observe, in the first place, that it cannot be proposed to extract the square root of the two members of an inequality, unless they are essentially positive, for otherwise imaginary expressions would be obtained.

But when we have 9<25; we deduce from it √9 or 3√25 or 5.

From a2>b2 we deduce a>b, if a and b express absolute numbers.

In like manner the inequality a2>(c-b) gives a>c-b if we suppose c greater than b, and a >b—c when we suppose that bis greater than c.

When the two members of an inequality are composed of additive and subtractive terms, care should be taken to write for the square root of each member, a polynomial in which the subtractions will be possible.

Problems.

106. Problem 7. Two merchants each sold the same kind of stuff; the second sold 3 yards more of it than the first, and together, they receive 35 crowns. The first said to the second, I would have received 24 crowns for your stuff; the other replied, and I would have received 12 crowns for yours. How many yards did each of them sell?

Problem 8. Amerchant has two notes out, one for 6240 francs payable in 8 months, the other 7632 francs payable in 9 months. He draws in these two notes, and in place of them gives one for 14256 francs payable in one year. What is the rate of interest? (Ans. 10.33 per cent.)

Problem 9. A widow possessed 13000 francs, which she divided into two parts, and placed them at interest, in such a manner, that the incomes from them were equal. If she had put out the first portion at the same rate as the second, she would have drawn for this part 360 francs interest, and if she had placed

the second out at the same rate as the first, she would have drawn for it 490 francs interest. What were the two rates of interest?

N. B. The equation of this problem can be resolved in a more simple way than by the general method.

Problem 10. Find two rectangles, knowing the sum (q) of their surfaces, the sum (a) of their bases, and also the surfaces (p and p') when to the base of each of them we give the altitude of the other.

Resolve this problem, and discuss it.

{

Ans. Base of the first. x=

a[2p+q±√ q2 —4pp'
2(p+p'+q)

Problem 11. Divide two numbers a, and b, each into two parts, such that the product of one part of a by a part of b, may be equal to a given number p, and that the product of the remaining parts of a and b may be equal to another given number p'.

Resolve this problem and discuss it.

Problem 12. Find a number, such that its square may be to the product of the differences between this number and two other numbers, a and b, in the ratio of p: q.

This last problem is recommended, not only because its discussion presents new applications of the rules for inequalities, but because the formulas obtained from it contain implicitly the solutions of a great many analogous questions, the enunciations of which only differ as to the sense of certain conditions.

Questions of Maximum and Minimum. Properties of Trinomials of the Second Degree.

107. There is a certain class of problems, frequently met with in the application of algebra to geometry, which refer to the theory of equations of the second degree. The object of these questions is to determine the greatest or least values that the result of certain arithmetical operations effected upon numbers may receive.

Suppose the question proposed is: To divide a given number, 2a, into two parts, the product of which shall be the greatest possible, or a maximum.

Denote one of the parts by x, the other will be 2a-x, and their product (2a-x). By giving a different values, this product will pass through different states of magnitude, and it is required to assign that value to x, which will render this product the greatest possible.

Denote this greatest product, which is unknown, by y; we will have from the enunciation the equation

x(2u-x)=y.

Proceeding as though y was known, we find the value of

x to be

x=a±√a2-y.

Now this result shows that the two values of x cannot be real, unless we have y<a2, or at most y=a2; whence we may conclude that the greatest value that y, or the product of the two parts, can receive, is a2. But by making y=a2, we find x=a.

Therefore, to obtain the greatest product, it is necessary to divide the given number into two equal parts, and the maximum obtained is the square of half of the number, a result which has already been obtained by another method. (No. 100). More simple solution. Let 2x be the difference between the two parts; since their sum is expressed by 2a, the greater will 2a + 2x be represented by

2

(No. 4), or a +x, the less by a-x,

and the equation will be (x+a) (x—a)=y; or, performing the operation indicated, a2-x=y; whence,

x= ±√ a2 —y.

In order that this value of x may be real, it is necessary that the value of y should not be greater than a2, and making y=a2, we have x=0, which proves that the two parts must be equal to

each other.

This solution has the advantage of leading to an equation of the second degree, involving two terms.

108. N. B. In the above equations, x(2a—x)=y, and . . (a+x) (a—x)=y, the quantity x is called a variable, and the expression x(2a—x), or (a+x) (a—x), is said to be a functi of the variable. This function, represented by y, is it another variable, the value of which depends upon that at buted to the first. For this reason, analysts sometimes

(

the first the independent variable, whilst the second, or y, receives values depending upon those attributed to x.

By resolving the two equations (2a—x)=y, and (a+x) (a−x)=y, with reference to x, which gives x=a±√ a2 −y,

we can consider y as an independent variable, and x as a certain function of this variable.

109. Suppose that it is required to divide a number 2a, into two parts, such that the sum of their square roots shall be a maxi

mum.

Call one of these parts x2, the other will be, 2a-x2, and the expression for the sum of their square roots will be x+ √2a-x; it is required to determine the maximum of this expression.

Suppose x+ √2a-x2=y.

In order to resolve this equation, it is necessary to get clear of the radical. By transposing the term x, into the second member, we have

whence, by squaring,

√2a-x2=y-X,
2a-x2=y2-2yx+x2,

or, arranging it with reference to x, 2x2-2xy=2a-y2,

In order that these two values of x may be real, it is necessary that y2 should not exceed 4a; hence 2a is the greatest value that y can receive.

But by making y=2√a we find x= √a, whence x2=a, and

2a-x2-a.

Therefore, the given number 2a must be divided into two equal parts, in order that the sum of the square roots of these parts may be a maximum. This maximum is moreover equal to 2 √a.

For example, let 72, be the proposed number; we have 72= 36+36; whence √36+ √36=12; this is the maximum of all the values that can be obtained for the sum of the square roots of the two parts of 72.