Recall that if S is a subset of n-dimensional space and P is a point of S we say that P is a point in the interior of S or a point inside S if there is some (small) positive number r such that every point of n-dimensional space within distance r of P is a point of S.
Recall that a function f of n variables is differentiable at a point inside its domain if it admits first order approximation by a linear function near the given point.
Theorem: If a function f of n variables has an extreme value for the subset S of its domain at a point P of S that is a point inside the domain of f where f is differentiable, then the gradient vector \nabla f (P) of f at P must be perpendicular to the tangent vector at P of every differentiably parameterized curve lying in S and passing through P.
Proof. Let G(t) be a differentiably parameterized curve contained in S and passing through P when t = a. Since S is contained in the domain of f, the function h(t) = f(G(t)) is defined for all values of t for which G(t) is defined, and since f is differentiable at P = G(a), the function h is differentiable at a. In fact, the ``chain rule'' tells us that
Since f has an extreme value relative to the set S at the point P and each G(t) is in S, it follows that h, a function of one variable, has a local extreme value at t = a, and, therefore, that h'(a) = 0. Consequently, \nabla f(P) is perpendicular to the tangent vector G'(a) of the curve at P.
Corollary 1. If f a function of n variables has an extreme value for the subset S of its domain at a point P of S that is a point inside S where f is differentiable, then the gradient vector \nabla f(P) must be the zero vector.
Proof. If P is a point inside S then every sufficiently short line segment passing through P must be perpendicular to \nabla f(P), which means that every vector must be perpendicular to \nabla f(P).
Corollary 2. If f is a function of n variables has an extreme value on the subset S = { g = 0} of its domain, where g is a differentiable function, at a point P of S, then the gradient \nabla f(P) of f and the gradient \nabla g(P) of g must be parallel vectors.
Proof. The statement is formally true, but probably useless if \nabla g(P) = 0. We assume that \nabla g(P) is not the zero vector. In this case \nabla g is perpendicular to the tangent hyperplane (i.e., plane if n = 3 or line if n = 2) to S at P. Every unit vector in the tangent hyperplane is tangent to some small differentiably parameterized curve segment lying in S and passing through P. Hence, by the theorem, \nabla f(P) is also perpendicular to each such curve segment, and, hence, to the tangent hyperplane. Since a hyperplane has only one parallel class of normal vectors, \nabla f(P) and \nabla g(P) must be parallel.
Remark. The theorem is useful also in the case where f is a function of 3 variables and the constraint set S is a curve in space. Then the fact that P lies in S corresponds roughly to two equations for P and the orthogonality condition of the theorem provides, in non-degenerate situations an additional equation with the result that (usually) only finitely many such P are possible. (Among these are points that are maxima, minima, and those that are neither.) This is equivalent to the principle of ``Lagrange multipliers'' discussed in the text.