f(x, y), dx+ dy; and du must vary in reference to both x and y; therefore we have dx dx dy' dy2 efficients of u. The first is found by taking x alone variable, the third by taking y alone variable, and the second by differentiating u, first with respect to one of the quantities x and y as the variable, and the result with respect to the other as the variable. The higher differentials may be found in a similar manner. Find the second differential of u = e* y*. d u dx dy and = d2 u dy" 12ey; hence we have d'u= e* y* dx2+8ey3 dx dy + 12e y'dy'. 4 ey3, MAXIMA AND MINIMA FUNCTIONS OF TWO VARIABLES, 58. Let u = f(x, y), where x and y are two independent variables; but we are at liberty to assume that y (= x) is a function of x, provided that x is any arbitrary function of x. Now since y is a function of x, we cannot regard y and x as varying uniformly, as in No. 57; hence if dr is considered as constant, dy must be regarded as variable, and differentiating on this supposition, we get t du du du dy dx = d2 u t d2 u and dx2 d x2 dx dy . (1), dx + where the last term of (2) is found by considering as variable, and dx t du, t d'u, are the total differentials of u. Reasoning as in the case of maximum and minimum functions of one variable, it will be found that in case of either a maximum and minimum value of u, we must have case of a minimum, the exceptions being precisely the same as those mentioned in the case of a single variable (Art. 48). dy 0, we have by (1), being an arbitrary quantity, dx du dx dx = Equation (4) reduces (2) to = +2 + (5), dx dy' d x2 d x2 dx dy dx2 and the second member of (5) must, with the exception referred to, be negative in the case of a maximum, and positive in case of a minimum, not changing its sign whatever values may be attributed to x and y in the immediate vicinity of those which render u or f (x, y) a maximum or minimum. Put now the second member of (5) equal to zero, and Hence the second member of (5) will retain its sign unchanged if Consequently if u = f(x, y) be a maximum or minimum, we must d2 u are positive, u will be a minimum. EXAMPLES. 1. To find the maximum and minimum values of u = 41 2. The perimeter of a triangle is given equal to 2s; find the sides so that the area may be a maximum. Let x, y, and 2 s the area is = √ {s (s - y denote the three sides of the triangle; then x) (s− y) (x + y − s)}, which is to be a maximum. Its square will therefore be a maximum, and the logarithm of the square will also be a maximum; therefore u = log s + log (s − x) + lög (s − y) + log (x + y − s) = a maximum; Whence we get x = s, and 2 s y = = 0 = y = s; therefore the tri angle is equilateral, and the test shows the area is then a maximum. Hence dx = a minimum. y, which shows that 3/2 a3 = a 3/2. Hence the 2 a3 = xy, and consequently x = square whose side is x = d2 u dx d2 u dy = a3 α 3/2 2 = d2 u dx dy ( = 1 - 1 = 3; dx dy 4 aR 4. Inscribe in a given circle a polygon of n sides, having its area a maximum. Draw from the centre, O, of the given circle to the extremities of the successive sides of the polygon, the radii OA, OB, OC, etc., and denote the successive angles A OB, BOC, CO D, etc., by 01, 02, 03, etc., and the radius of the circle by r. Then the area of the isosceles triangle AO B = 72 sin 01, that of BOC = r2 sin 02, and so on; hence the entire area of the polygon will be r2 (sin 0, + sin 4 hence, rejecting the constant multiplier, we get u = sin 0, + sin 0, + sin 03 + . . + sin (2π du do1 Hence = cos 01 du d02 cos (2π − s), etc.; but these partial differential coefficients are severally = 0; hence and the polygon has, therefore, all its sides and angles equal, and it is, therefore, a regular polygon. 5. Among all rectangular prisms, to determine that which, having a given volume a3, shall have the least surface. Let x, y, z denote the three contiguous edges of the prism; then by the conditions which give x = = y = a, and thence z = a; therefore the prism is a cube. 6. Divide 24 into three parts, x, y, z, such that u = x y z3 may be a maximum. Ans. x = 4, y 8, and z = 12. 7. The surface of a rectangular parallelopiped is 2 a2; find when its volume is a maximum. Ans. x = y = z = a, or a cube whose side is a. 8. Inscribe the greatest parallelopiped in a given ellipsoid. Ans. If 2a, 2b, 2c, be the principal diameters of the ellipsoid, 2 a 26 3 2 c then √3, √3, √3 are the edges of the greatest 3 parallelopiped. 9. When is u = x2 + xy + y2 - 9x6y a maximum or minimum? 21, a minimum. 10. Show that the least polygon of a given number of sides which can be described about a given circle is a regular one. SINGULAR VALUES OF FUNCTIONS. 59. The value of an explicit function can generally be determined by performing the operations indicated, but it sometimes happens that such a value may be given to the variable as shall render the determination of the value of the function impossible in the ordinary manner. Thus, if the usual way. Such values are termed singular values, and they may be easily determined by the method of limits. For since (39) Now as x approaches to zero, the second member of the last equation 1 1.2.3 1 ; hence when x = O, we get u = 6 approaches to These singular values of functions may frequently be detected by the differential calculus in the following manner. Let v and z be the func 0 when v, we get by differentiation .(1). but when x = a, z = 0, therefore u = If the value xa makes the above process, gives u = d (dv) dz = O and dz = d2v d(dz) d'z Continue this process till one of the differential coefficients becomes finite, when x = a, and the value of the function will be determined. dv = n cos n x dx-n dx cos n x+n2x sin n x dx = n2 x sin nx dx; The singular value of u, when x = O, is therefore when x = = 0. |
