Ex. 2. Let there be only one unknown quantity x, for which the values au a?,... am have been found by m observations of equal weight. We have then to determine x so that (x-a1)i+(x-a2)2 + ... + (x-am)2 = M2 may be a minimum. Differentiating this expression and equating to zero the differential, because « is to be a minimum, we have tHX — d\ — <z2 — • • • — am — 0 3 «i + a» + ... + Am X = m Thus, assuming the validity of the method, the best value for x is the arithmetic mean of the values given by the observations. It was in fact by assuming this result as an axiom, that Gauss first demonstrated the validity of the method. But this is not self-evident, and is not properly axiomatic; and in fact no demonstration has yet been given perfectly free from difficulty, so far as I know. If the weights of the observations, instead of being equal, were gu git... g,„, the function of which the minimum value is to be determined would be Si {x - atf + ffi (x - otf + ... + gm (x-am)2; and the resulting value of x is ffi«i+ffa«2+ ••• + gm a,n ffl+ff2+ ••• +ffm Section 5.—On maxima and minima of functions of variables when all are not independent. 167.] A problem, which frequently occurs, is the determination of maxima and minima of functions of many variables, when certain relations between the variables are given, so that all those involved in the original functional equation are not independent. Thus suppose that we have to determine the singular values of M = v{x,y,z,...), (49) which is a function of n variables; and suppose besides that m equations connecting these variables are given, viz. In order to apply the method which has been explained in the preceding Section, it would be necessary to eliminate m variables between m +1 equations, by which means u would become a function of n — m variables, all of which would be independent of each other; and then forming the partial derived-functions ^7^' ^6^' ' *'ie num^er °^ wmcn 's n — m, and equating each to 0, there would be n — m equations, from which we could, theoretically at least, determine the n — m variables. This method however, though theoretically possible, is frequently attended with great difficulty on account of elimination; and, if the original expressions are symmetrical, the symmetry is destroyed by it; in which case it is better to proceed as follows: It is plain from what has been said, that as there are n — m variables entirely independent in their variations, we have n — m conditions to make; which will be equivalent to equating to 0 the n — m partial derived-functions with respect to these variables of r(x,y,z, ). Differentiating the functions in order, and remembering that Dm = 0, because u is to be a maximum or a minimum, we have dr' dx the meaning of which is, that x, y, z, do not vary independently of each other, but consistently with the conditions involved in the last m equations. Hence to eliminate dx, dy, dz, multiply these last equations severally by indeterminate quantities, \u A2, A3, A,„, and add to the first; and collecting the coefficients of dx, dy, dz, we have {©+'«.(£>+«•(£)+ •! I + =0. Which equation is subject to n conditions, viz. n — m, on account of n — m independent variables being involved, and m on account of m indeterminate multipliers Aj, A2, A3, A„ having been introduced. Let these conditions be, that the coefficient of each differential is equal to 0; therefore between which equations Ai, A2,... A„ are to be eliminated, and x,y,z, ... determined; and these will be the values corresponding to a maximum or a minimum value of r(x, y, z,...). The sign of the second differential coefficient will determine whether the particular value is a maximum or a minimum: but in most cases where this method is applicable, the form of the function at once decides whether it admits of a maximum or of a minimum. 168.] In the case in which only one equation is given connecting the n variables involved in the given function, whose maximum or minimum value is to be determined, the above results assume a particular form, by means of which the process is much simplified. Let u = r(x,y,z,...) (53) be the function of which the maximum or minimum value is to be determined; and suppose the variables to be subject to the relation expressed by the equation f(x,y,z,...) = c; (54) then differentiating (53) and (54), and putting D» = 0, because M is a maximum or a minimum, we have Price, Vol. I. N n x-a2 + b2 + c2, y~a2 + b2 + c2, "~ oa+62 + cs" The above is the solution of the problem, To find the shortest distance from the origin to a given plane. Ex. 2. To determine the plane triangle of given perimeter, and of maximum area. Let the area of the triangle = u, and the sides be x, y, z, and the given perimeter = 2 s. .-. 2a = x + y-\-z, and u2 = s (s—x) (s—y) (s—z); therefore differentiating the former equation, and taking the logarithmic differential of the latter, we have 0 = dx + dy + dz, 2 Dm dx dy dz = + —— + = 0; u s—x a — y s — z .-. s—x = 8—y = s—z; .•. x = y = z = ; and the area of the triangle = ^; that is, the triangle is equilateral. In precisely the same manner it may be shewn that of all figures of a given number of sides and of given perimeter, the equilateral and equiangular one is the greatest. Hence also it follows that, of all plane figures of given perimeter, the circle is that whose area is the greatest. For let 2 s be the given perimeter, and n be the required number of sides; .■. each side = ——: 8 7T and .-. the area = —cot-, But the last factor continually increases as - decreases, and attains a maximum when n = 00; in which case the polygon becomes a circle, which is therefore the greatest of all plane figures under a given perimeter. Ex. 3. To divide a number a into three parts x, y, z, so that a?" yn z? may be a maximum. u = xmynzp, .•. logw = mlogx + wlogy + p \ogz; |