he would have received 24 crowns; had the first sold the same as the second, he would have received 124 crowns. How many ells

did each sell?

First Ans. The first sold 15 ells, the second 18.

Second Ans. The first sold 5 ells, the second 8. Problem 8. A merchant owes 6240l. payable in 8 months, and 76321. payable in 9 months. To pay these, he gives a bill of What is the rate of interest?

14,2561. payable in a year.

Ans. 10,33 per cent.

Problem 9. A man has 13,000l. which he divides into two parts, and places them out at interest so as to derive the same income from each. Had the first sum been placed at the same rate of interest as the second, it would have yielded 360l. per annum; had the second been placed at the same interest as the first, it would have yielded 490l. per annum. What were the rates of interest? Ans. 7 and 6 per cent.

N. B. The equation of this problem may be solved more simply than by the general method.

Problem 10. The sum of the areas of two rectangles is q, the sum of their bases is a, and p and p' are the areas of two rectangles, having respectively the base of the first, with the altitude of the second, and the base of the second with the altitude of the first. Required the resolution and discussion of this problem.

The base of the first is

a [2p+q±√√√ q2 — 4pp'] 2 (p + p' + q)

Problem 11. Required the solution and discussion of the fol lowing problem: To divide each of two numbers a and b into two parts, so that the product of one part of a by one part of b may be a given number p, and the product of the remaining part of a by the remaining part of b may be another given number p'.

Problem 12. To find a number such that its square divided by the product of the differences of this number and two given numbers a and b, may be equal to ; required the resolution and discussion

of this problem.

P

q

Questions concerning Maxima and Minima. Properties of Trinomials of the Second Degree.

107. There is a certain class of problems which relate to the theory of equations of the second degree, and which we often meet with in the Application of Algebra to Geometry. The object of these questions is to determine the greatest or least value which the result of certain arithmetical operations, performed upon numbers, is susceptible of.

Let it be proposed, as the first question, to divide a given number 2 a into two parts, whose product shall be the greatest possible,

or a maximum.

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

Let us designate by y this greatest product, whose value is at present unknown; we shall have, according to the enunciation, the equation

x (2 a—x) = y.

First regarding y as a known quantity, and deducing from this equation the value of x, we find

x = a ± √ a2 — y•

Now it is evident from this result, that the two values of x can be real only while we have y <a2, or at most y = a2; whence we may conclude that the greatest value which can be given to y, or the product of the two parts, is a2. But if we make y = a2, there hence results x = a.

So that, in order to obtain the greatest product, we must divide the given number 2 a into two equal parts, and the maximum which we obtain is the square of half of the number, a result at which we have arrived by another method (100).

A more simple solution. Let 2 x be the difference between the two parts; since their sum is already expressed by 2 a, the greatest of these parts will be (4) respresented by

2a + 2x, or a + x,

2

the least by a

x, and we shall have for the equation,

(a

(a + x) (α — x) = y;

or, performing the calculations, a2

x2 =

y; whence

In order that this value of a may be real, it is necessary that y should be at most equal to a2; and by making ya2; we obtain x = O, which proves that the two parts ought to be equal.

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

108. N. B. In the equations

x (2 a − x) = y and (a + x) (a — x) = y,

established above, the quantity x is what we call a variable, and the expression x (2 a − x), or (a + x) (a — x), is called a certain function of the variable. This function, represented by y, is itself another variable, whose value depends on that which we give to the first. It is for this reason that analysts sometimes designate the former by the naine of independent variable, while the latter, or y, receives values dependent upon those which we give to x. Resolving the two equations.

x (2 a — x) = y and (a + x) (a — x) = y,

we may in turn regard y as an independent variable, and x as a certain function of that variable.

109. Let it be proposed, as a second question, to divide a number 2 a into two such parts that the sum of the square roots of these two parts may be a maximum.

Let us call a one of the parts, 2 a — 2 will be the other part, and the sum of their square roots will have for its expression

x + √2a — x2;

it is this expression whose maximum is to be determined.

To resolve this equation, we must free it from the radical. We have first, by transposing the term a to the second member,

In order that the two values of x may be real, y2 must be at least equal to 4a; then 2a is the greatest value which y can receive.

But if we make y = 2a, there results xa, whence we deduce x2= a, and 2 a x2 = a.

Thus the given number 2 a must be divided into two equal parts, in order that the sum of the square roots of these two parts may be a maximum. This maximum is moreover equal to 2a. For example, let 72 be the proposed number; we have

whence

72 36 36;

√36 + √36 = 12;

this is the maximum of the values which can be obtained as the

sum of the square roots of the two parts of 72.

64 = 8,

10+ a fraction.

For let us decompose 72 into 64 +8; we have 82+ a fraction; whence 64+/8

Again, let there be 72 49 +23; we have

then

/497, 23 4+ a fraction;

√49+ √/23 = 11+ a fraction.

Let us consider, for the third example, the expression

which it is required to render a minimum (m being supposed >n).

Let us put

Now, in order that the two values of x, corresponding to a value

of y, may be real, it

is evident that (m2n2)2 y2, must be at

least equal to 4 m2 n2, and consequently that y must be at least

110. These examples are sufficient to show the steps which must be followed in the resolution of questions of this kind.

After having formed the algebraic expression of the quantity susceptible of becoming either a maximum or a minimum, we make it equal to any letter whatever y. If the equation which is thus obtained, is of the second degree in terms of x, (x designating the variable quantity which enters into the algebraic expression,) we resolve it with reference to x; then make equal to zero the quantity under the radical, and obtain from this last equation a value of y, which then represents the maximum or minimum sought. Lastly, substituting this value of y in the expression for x, we have the value of this last variable, which will satisfy the enunciation.

N. B. If it should happen that the quantity under the radical remains essentially positive, whatever be the value of y, we should conclude that the proposed expression can pass through all possible states of magnitude; in other words, that it would have infinity for a maximum and zero for a minimum.

Let there be, for a new example, the expression