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ON MAXIMA AND MINIMA.

DEF. * Of all straight lines, angles, or areas, subject to given conditions, the maximum is that which is the greatest, the minimum is that which is the least.

PROP. 68. To divide a given straight line into two parts, so that the sum of the squares of the segments (1) shall be given: (2) shall be a minimum.

* Though the above definition may suit as far as the point, line, and circle are concerned, still the following is a more general one :—

When any variable magnitude, linear or otherwise, subject to given conditions, passes gradually from an increasing to a decreasing state, or vice-versa, it has, in the former case, gone through a maximum value, and in the latter through a minimum.

The dotted lines MP, mp, in the diagram, are, respectively, maxima

and minima.

Let AB be the given straight line: it is required (1) to divide it into segments, the sum of the squares of which shall be equal to a given square X.

At B, with the straight line AB, make the angle ABL equal to half a right angle. With A as centre, and a radius equal to a side of the square X, describe a circle. Let this circle cut the line BL in K; join AK; and draw KD perpendicular to AB: AD, BD are the required seg

ments.

In the triangle BDK, the angle B is half a right angle, and the angle D is a right angle; therefore the angle DKB is half a right angle (I. 32); and, therefore, the sides DK, DB are equal (I. 6). Again, since the angle ADK is right, (I. 47) AK2= AD2 + DK2 = AD2 + BD2. But the square on AK is equal to the square, X, by construction; therefore AD2 + BD is equal to the given square X

(2) *To divide AB in D so that the sum of the of AD, BD shall be a minimum.

squares

*The student can solve (2), independently, by means of (II. 9).

The construction being the same, we see that the sum of the squares of any two segments, AD, BD, varies as the square of a line drawn from A to the line BL. Hence, when the sum of the squares of AD, BD is a minimum, the line AK is a minimum, which is (I. 19) the case when AK is perpendicular to BL. The point, D, will then become (I. 32, 26) the bisection of AB, that is to say, the segments are equal.

PROP. 69. To divide a given straight line into two parts, so that the rectangle under the segments (1) shall be given: (2) shall be a maximum.

Let AB be the given straight line: it is required (1) to divide it into segments, the rectangle under which shall be equal to a given square.

Upon AB describe a semicircle, and at any point, E, in the base, erect EF perpendicular, and equal to a side of the given square. Through F draw FCC' parallel to AB, and cutting the circumference of the semicircle in C, C;

draw CD perpendicular to AB: AD, BD are the required segments.

For, since the angle ACB is right, being in a semicircle, AD. BD= CD2 (10); but CD, FE are equal, being the opposite sides of a rectangle; therefore

AD.BD=FE2 – the given square.

=

(2) *To divide AB in D so that AD. BD shall be a maximum.

From the first part, we see that AD. DB varies as the square of CD. Hence, the rectangle will be a maximum when the square is a maximum; that is, when CD is a maximum; which, clearly, is the case when C is the bisection of the semicircular arc. The point, D, is, then, obviously, the bisection of the line AB.

Note. (1) may evidently be stated thus:

Given the rectangle under two lines, and the sum of the lines: to find them.

(2) clearly shews that :

Of all rectangles of a given perimeter, the square has the maximum area.

EXAMPLE.

Through a fixed point, within or without a given circle, draw a straight line cutting the circle, so that the sum of the intercepted segments shall be equal to a given length. When would the problem be impossible?

[Since the rectangle under the segments of the secant is always constant (III. 35, 36), according as the point is within or without the circle, the problem is therefore reduced to the above. See Note (1).]

*

This, also, can be solved, independently, by means of (II. 5).

PROP. 70. Of all rectangles of a given area, the square has the minimum perimeter.

E.

G

Let DEFG be any rectangle of the given area. Produce DE to H, so that EH is equal to EF. Upon DH, as diameter, describe a semicircle. Let EF, produced, meet the semicircle in C. Join DC, HC, OC, O being the middle point of DH.

Since the angle DCH is right, being in a semicircle, and CE is perpendicular to DH, then (10) DE. EH =CE2. But EH, EF are equal, therefore the rectangle DEFG is equal to the square on EC.

Now, the perimeter of the rectangle is, obviously, double the line DH, or four times the line OC. But, in the right-angled triangle OCE, the hypotenuse OC is greater than the side CE (I. 19); therefore four times the line OC is greater than four times the line CE, in other words, the perimeter of the rectangle is greater than the perimeter of the square.