or a, x, y, z, b are in geometric progression. Let log u = v, then proceeding to the second differentials we get, on substituting for x, y, ≈ the values na, na, n3a, dv dy dz = 1 1 a2 n2 (1 + n)2 d2v = 0; a2 n1 (1 + n)2 ' dx dx so that Lagrange's condition becomes a2 n10 (1 + n)* anto (1 + n) a and y being connected by the equation 1 = (1 + p2) x2 + 2pq xy + (1 + q2) y2. Differentiating these two equations, 0 = (rx + sy) d x + (s x + ty) dy, 0 = {(1 + p2) x + pqy} dx + {pqx + (1 + q2) y} dy. Multiply the second of these equations by an indeterminate quantity A, add it to the first and equate to zero the coefficients of the differentials: this gives λ {(1 + p2) x + pqy} + rx + sy = 0, \ {pqx + (1 + q2) y} + sx + ty = 0. Multiply these equations by a, y respectively and add, k then, by the original equations λ + − = 0. Substituting this value of λ in the preceding equations, and grouping together the terms multiplied by the same variable, Multiplying these together so as to eliminate x and y, we find k (1 + p2 + q3) · − = {(1 + q3) r − 2 p q s + (1 + p3) t } + r t − s2 = 0, k a quadratic equation in whence a maximum and a mini ρ mum value may be found. This is the equation for determining the radii of maximum and minimum curvature in a curved surface. (6) Let u=ay (c≈) = bz (a − x) = cx (b − y). Then a = a, y = 1b, ≈ = 1c, give u =abc, a maximum. (7) Let u = a cos2x + b cos2y; ', x and y being subject to the condition y Differentiating, we have - x= = 4 Απ 0 = a cos x sin x + b cos y sin y, or 0 = a sin 2x + b sin 2y. From the equation of condition The upper sign giving a maximum and the lower a minimum. (8) Let u = cos x cos y cos x, with the condition + y + z = π. By taking the logarithmic differential, and using an indeterminate multiplier, we easily find = 4, a maximum. x = y=x= |, and u = (9) Find the maximum value of u = al + bm + cn, l, m, n being variable and subject to the condition We easily find by the use of an indeterminate multiplier that and therefore u = (a2 + b2 + c2) §. This is the solution of the problem, "To find the position of the plane on which the sum of the projections of any number of planes is a maximum :" l, m, n are here the cosines of the angles which the plane of projection makes with the co-ordinate planes. (10) Find the maximum value of u = (x + 1) (y + 1) (≈ + 1), x, y, z being subject to the condition, da loga + dy log b + d z log c = 0. Whence, by using an indeterminate multiplier A, This is the solution of the problem: "If a, b, c be the prime factors of a number 4, to find how many times each factor must enter into it, that it may have the greatest number of divisors." Waring, Medit. Algeb. p. 344. (11) To find the rectangular parallelopiped which shall contain a given volume under the least surface. If x, y, z be the edges of the parallelopiped, and if a3 be the volume of a cube to which it is equal, then by the condition of the minimum we easily find x = y = ≈ = A, so that the surface equals 6a, a minimum. (12) To inscribe the greatest rectangular parallelopiped in a given ellipsoid. Let the equation to the ellipsoid be and let x, y, ≈ be the half edges of the parallelopiped, then u = 8xyz is to be a maximum, x, y, z being subject to the condition (13) To find the triangle of least perimeter which can be inscribed in a given triangle. Let ABC (fig. 9) be the given triangle, a, b, c the sides, A, B, C the angles, and DEF, being the inscribed triangle, let CD = x, AE = y, BF = x. Then if u be the perimeter, {y2 + (c − 2)2 - 2y (c − ≈) cos A} } From these equations it appears that FEA = DEC, EDC = BDF and BFD = AFE. It is shewn by Geometry that if lines be drawn joining the points where the perpendiculars from the angles meet the sides, each intersecting pair makes equal angles with the side in which they meet; consequently the triangle formed by these lines is the triangle of least perimeter which can be inscribed in the given triangle. See Cambridge Mathematical Journal, Vol. I. p. 157. (14) To find the least distance between two straight lines in space. |
