\HHT Method of Lagrange Multipliers h\\b\w \bMethod of Lagrange Multipliers.b\ \bProblem.b\ For given function f(x) where x=(x\[1 \ ,x\[2 \ ,...,x\[n \ ) - n variables, find an extrema if x belongs to domain H which is set by one or more constraints: g(x) = 0 or g\[1 \ (x) = 0 g\[2 \ (x) = 0 ... g\[m \ (x) = 0. \bNote.b\ Summation rule x\[1 \ y\[1 \ + x\[2 \ y\[2 \ + ... = x\[i \ y\[i \ is assumed below. \bIdea.b\ Let's start from case m=1. As for three dimesional case, let's call domain H and contour f(x)=const as "hypersurface". As in two-dimesional case n=2, suppose that contour of f is "tangent" to the hypersurface H in point of extrema x=e. We need to develop more precisely what does this mean "tangent". In neighbourhood of p, an arbitrary point of hypersurface H, equation of hypersurface g can be written as: Gx=0 if to choose the point p as an origin, denote j-th partilal derivative of g with Gj, and shortcut inner product G\[j \ x\[j \ with Gx. The last equation means that H is approximated with S, where S is vector subspace of IR\]n \ , S is orthogonal to G, and dim S = n-1. Contour's C equation f(x)-const=0 can be written in similar form Fx=0, where F = (F\[1 \ , F\[2 \ , ... , F\[n \ ), F\[j \ is partial derivative f by x\[j \ , and contour C is approximated with subspace T which is orthogonal to F. Now, we can precisely say that T and S are tangent iff F and G are parallel, or if G is not vanished, there exist number L that (*) F - L*G = 0, or grad(f-Lg) vanishes in point p. But, why T and S must be tangent? At this point, it can be seen that developed language allows to formulate condition of extrmality without geometrical language which brought us so far. Indeed, for p=e, for each x from S, Fx must be zero. Otherwise, f(x)-f(0) and f(-x) - f(0) have different sign and p is not an extremum. This means that F and S are orthogonal which implies (*). (As F and G are vectors in n-dimensional space which both are orthogonal to (n-1)-dimensional subspace.) h\