du Now if the expression for an be multiplied by (x + 2)" and divided by (x + 3)', which are both essentially positive quantities, the result will be equal to v = x (w + 2). . whence x = , and u = *, a maximum, This is the solution of the problem, To find the height at which a light should be placed so that a small plane surface at a given horizontal distance shall receive the greatest illumination from it. This is a solution of the problem, To find the magnitude of the body which must be interposed between two others so that the velocity communicated from the one to the other shall be a maximum. This is a solution of the problem, To find in what direction a body must be projected with a given velocity that its range on a given plane may be the greatest possible. (13) (sin mx) (sin x)? The values of x derived from m tan x = tan mx, make u a maximum. The values of x derived from sin ma = 0, make u a minimum. The values of a derived from sin x = 0, make u a maximum and equal to m2. Airy's Tracts, p. 328. (ax x?)! 2 at + af. Here du = £ ¢ (a – a)-, hence, multiplying by hich is essentially positive, we have v=c(x – a) = 0; if x = a, ao = c. Therefore if c be positive x = a makes u a minimum ; and if c be negative, a maximum. (19) Let u = b + c (x – a) change of sign, it appears that x = a which makes dc u = (1 + x3) (7 – x)?. X = 0 gives u a minimum. (21) To divide a number a into a number of equal parts such that their continued product shall be a maximum. Let æ be the number of parts, then is one of the parts, and () is the continued product, which is to be a maximum. Taking the logarithmic differential, we have log a – log x – 1 = 0. (23) To find the point in the straight line AD (fig. 2), at which BC subtends the greatest angle; ABC being perpendicular to AD. If the angle be a maximum its tangent is also a maximum. ab + x 2 ? a na im. (24) Through a point M (fig. 3) within the angle BAC draw the line PQ so that the triangle PAQ shall be a minimum. Draw MN parallel to AQ, and let AN = a, MN = b, AP = x. Then x = 2a makes PAQ a minimum, and PQ is bisected in M. (25) Given the length of the arc of a circle, find the angle which it must subtend at the centre in order that the corresponding segment may be a maximum. Let a be the half-length of the arc, x the radius of the circle ; then – is the half-angle of the segment. COS a maximum. If u be the segment, u = ax – xo sin cos. is a maximum. en in This last equation might be equally satisfied analytically by a value of between 7 and 3*, but such an angle is excluded by the geometry of the problem. A geometrical solution of this problem is given in the Mathematical Collections of Pappus, Book V. Theor. 16. (26) AC (fig. 4) and BD being parallel, it is required to draw from C a line CXY, such that the sum of the triangles ACX and BXY shall be a minimum. If AC = a, AB = b, AX = X, it is easily seen that the area of the triangle ACX is proportional to ax, and that of o that we have Whence we find me – b? = 0, or x = 64, which determines the line CXY. Vincent Viviani, Geometrica Divinatio, p. 152. |