Extreme Values of Functions of Several Variables

Math 214 Handout

March 16, 2001

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 ∇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

h′(a) = ∇f(P) · G′(a) .

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, ∇f(P) is perpendicular to the tangent vector G′(a) of the curve at P.

Corollary 1. 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 S where f is differentiable, then the gradient vector ∇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 ∇f(P), which means that every vector must be perpendicular to ∇f(P).

Corollary 2. If a function f of n variables has an extreme value for the subset S = {g = 0} of its domain at a point P of S where f and g are differentiable functions, then the gradient ∇f(P) of f and the gradient ∇g(P) of g must be parallel vectors.

Proof. The statement is formally true, but probably useless if ∇g(P) = 0. We assume that ∇g(P) is not the zero vector. In this case ∇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, ∇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, ∇f(P) and ∇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.

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