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aid only of quadratic equations: we shall here present a specimen of such solutions, following the example of Bourdon in his "Algèbre

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1. Divide a given number 2a into two parts, such that the product of those parts may be the greatest possible.

Let x be one part, then 2a present the product of these x (2 a — x) = y.

a is the other; let also y rewhen greatest: then we have

.. 2 ax - a
x2

= y, .. x2 2ax=

y.

Solving this quadratic, as if y were known, we have

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Now the greatest value that y can have consistently with a real value of x, is evidently y = a2. This, therefore, is the maximum value of the product 2ax-x2, that is to say, we must have 2ax - x2 = a2, or (1), x = a ± √ (a2 — a3) : Consequently, the product of the two equal parts of a number is always greater than the product of any two unequal parts of it.

= a.

2. Divide a given number 2 a into two parts, such that the sum of their squares shall be less than the sum of the squares of any other two parts into which that number may be

divided.

Let the parts be x and 2a

, and put y for the sum of the squares when this sum is a minimum: we then have

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y

Now the least value that y can have consistently with the reality of a, is evidently such as to render 2 a2 = 0, because for a less value than this the expression under the radi

cal sign would be negative, and therefore a would be imaginary, consequently y 2a; therefore (1),

=

x = a± √ (a3 — a3) = a,

so that, as in the former problem, the number must be divided into equal parts.

3. Divide a number 2 a into two parts, such that the sum of the quotients of each part by the other shall be a minimum.

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Equating the expression under the radical to zero, we have

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so that as before the number must be divided in two equal parts; the sum of the quotients being 2.

4. It is required to determine whether the expression 4x2 + 1x

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6 (2x+1)

3

-admits of a maximum or a minimum.

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Now whatever real value be given to y, the quantity under the radical sign will always be positive; so that x will continue real, however small or however great y be taken; therefore there is no limitation as to the smallness or largeness of y; that is, there is no maximum or minimum value of the proposed expression.

IV. To find the number of shot in a pile, whether square, triangular, or rectangular.

In arsenals and fortresses, cannon balls and shells are arranged in piles with either square, triangular, or rectangular bases.

A square pile is formed by a succession of horizontal courses one above another, each course forming a square, the pile being crowned by a single shot, the sides of the successive courses decreasing also by a single shot in regular order from bottom to top.

A triangular pile is made up of horizontal triangular courses, the sides of the successive courses decreasing by a single shot, and the pile terminating, as in the square pile, by one shot at top.

In both these piles the sides or faces are equilateral triangles, the shot in each face forming an arithmetical progression, the first of which is 1, and the last the number of shot in the bottom row.

If to one of the sloping faces of the square pile a new and equal face of shot be added, the whole pile will of course terminate in two shot; if another face be added it will terminate in three shot, and so on; and in this way the rectangular pile may be regarded as formed; the number of triangular faces of shot added to the square pile being thus always one less than the number of shot in the top row or edge; or, which is the same thing, this number is equal to the difference between the number of shot in the longer side and the number in the shorter side of the rectangular base.

1. How many shot are there in a square pile which has 8 shot in the bottom row?

Commencing at the top shot, and adding together the several courses, we have

1 + 22 + 32 + 42 + 52 + 62 + 72 + 82 =

or by the formula at page 145*

204,

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2. How many shot are there in an oblong or rectangular pile, the number of shot in one side of the base being 16, and the number in the other side 7?

By the above formula the number of shot in the square pile, the side of whose base has 7 shot, is

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The number of shot in a face of this pile is 1 + 2 + 3 + 4 + 5+ 6+ 7 = 28; and as 9 such faces have been added to the square pile, 28 × 9 = 252 shot have been added to that pile to form the rectangular pile.

.. 140 + 252 = 392, the number of shot in the pile.

It is plain that, as in this example, the number of shot in an oblong pile will always be obtained by adding to the square pile, the side of whose base is n-the shortest side of the oblong base—the arithmetical progression 1 + 2 + 3 ... +n multiplied by m- n, m being the longest side of the oblong base.

The general formula for the shot in an oblong or rectan gular pile is therefore

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* The truth of this general expression for the sum of the squares of the natural numbers, is established at Article V. of this Appendix.

3. How many shot are there in an oblong pile, the shorter side of the base of which contains 15, and the longer side 35?

Here n = 15, and m = 35.

15 x 16 x 91

shot.

6

= 5 × 8 × 91 3640, the number of

4. How many shot are there in a triangular pile, of which each side of the base contains 8 shot?

It is obvious that, commencing at the bottom course, the shot in the several courses are given by the following arithmetical progressions:

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It is plain that the n layers or courses in a triangular pile is always the sum of n terms of the series.

1 + 3 + 6 + 10 + 15 + 21 + 28 + &c.;

or, taking the terms in pairs, if there be an even number of

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* That the sum of any two consecutive terms in the series of triangular numbers as these numbers are, for obvious reasons, called-is always a square number, may be proved in general terms, as follows:The nth triangular number being the sum of the arithmetical series

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