Multivariable Calculus 19 | Examples for Local Extrema

TL;DR

A critical point x0 for a local extremum must satisfy ∇f(x0)=0.

Briefing Cornell Notes

Briefing

Local extrema in multivariable calculus hinge on two tests: the gradient must vanish at a critical point, and the Hessian matrix must have the right definiteness. For a twice continuously differentiable function, a critical point x0 where ∇f(x0)=0 is only a candidate; then the Hessian Hf(x0) decides what happens. If Hf(x0) is positive definite, the point is an isolated local minimum; if Hf(x0) is negative definite, it is an isolated local maximum. When the Hessian is indefinite, no local extremum exists—often the point becomes a saddle.

The transcript first revisits an eigenvalue-based way to check definiteness for symmetric matrices: a symmetric matrix is positive definite exactly when all its eigenvalues are strictly positive, and negative definite exactly when all its eigenvalues are strictly negative. This turns the Hessian test into a sign-check problem on real eigenvalues (symmetric matrices have real eigenvalues). The first worked example uses a two-variable quadratic function f(x1,x2)=x1^2+3x2^2. Its gradient is (2x1, 6x2), so the only solution to ∇f=0 is (0,0). The Hessian is the constant diagonal matrix [[2,0],[0,6]]. Because both diagonal entries (and thus both eigenvalues) are positive, the Hessian is positive definite. That definiteness guarantees an isolated local minimum at the origin, and the quadratic form also makes it the global minimum.

The second example flips the sign of the x2^2 term: f(x1,x2)=x1^2−4x2^2. The gradient becomes (2x1, −8x2), again yielding only one critical point at (0,0). The Hessian is [[2,0],[0,−8]], also diagonal, but now the eigenvalues have opposite signs: one positive and one negative. That mixed-sign pattern means the Hessian is indefinite. Indefiniteness rules out local extrema at the origin, and the geometric intuition matches: the surface curves upward in one direction (like a valley) and downward in the perpendicular direction (like a hill). The point is therefore a saddle point—looking like a maximum from one viewpoint and a minimum from another.

A key takeaway is that the Hessian can both confirm and deny local extrema. In these quadratic cases, eigenvalues are easy because the Hessian is diagonal. But when Hessians grow large, eigenvalue computation can become burdensome, motivating alternative definiteness tests. The transcript points ahead to Sylvester’s Criterion as a method that avoids eigenvalues and will be discussed next.

Cornell Notes

Local extrema in multivariable calculus come from combining two conditions: critical points where ∇f(x0)=0, and Hessian definiteness at x0. For symmetric Hessians, definiteness can be checked via eigenvalues: all positive eigenvalues imply a positive definite Hessian and an isolated local minimum; all negative eigenvalues imply a negative definite Hessian and an isolated local maximum. If the Hessian is indefinite (eigenvalues of mixed signs), the critical point is a saddle and there is no local extremum. Two quadratic examples illustrate both outcomes: x1^2+3x2^2 gives a positive definite Hessian and a minimum at (0,0), while x1^2−4x2^2 gives an indefinite Hessian and a saddle at (0,0).

Why must ∇f(x0)=0 before the Hessian test matters?

A point x0 is only a candidate for a local extremum if the gradient vanishes there. For continuously differentiable functions, ∇f(x0)=0 is a necessary condition for a local maximum or minimum. After finding such critical points, the Hessian Hf(x0) is used to determine whether the candidate is actually a minimum, maximum, or neither.

How does eigenvalue information determine whether a symmetric Hessian is positive definite or negative definite?

For a symmetric matrix, all eigenvalues are real. The Hessian is positive definite exactly when every eigenvalue is strictly greater than 0; it is negative definite exactly when every eigenvalue is strictly less than 0. In the examples, the Hessians are diagonal, so the eigenvalues are just the diagonal entries, making the sign check immediate.

In f(x1,x2)=x1^2+3x2^2, how are the critical point and the type of extremum determined?

The gradient is (2x1, 6x2). Setting it to zero forces x1=0 and x2=0, so the only critical point is (0,0). The Hessian is [[2,0],[0,6]], whose eigenvalues are 2 and 6—both positive—so the Hessian is positive definite. That guarantees an isolated local minimum at (0,0), and the quadratic form also makes it the global minimum.

In f(x1,x2)=x1^2−4x2^2, why does the origin become a saddle instead of a minimum or maximum?

The gradient is (2x1, −8x2), so ∇f=0 again only at (0,0). The Hessian is [[2,0],[0,−8]], with eigenvalues 2 (positive) and −8 (negative). Mixed signs mean the Hessian is indefinite. Indefiniteness rules out local extrema, and the geometry matches: the surface curves upward in the x1 direction but downward in the x2 direction.

What practical limitation appears when Hessians are large, and what alternative is hinted?

Eigenvalue computation can be difficult for large Hessian matrices. The transcript notes an alternative method—Sylvester’s Criterion—that can test positive/negative definiteness without explicitly finding eigenvalues.

Review Questions

For a twice continuously differentiable function, what two conditions are needed at x0 to conclude an isolated local minimum or maximum?
How does an indefinite Hessian (mixed-sign eigenvalues) change the conclusion about local extrema at a critical point?
In the example f(x1,x2)=x1^2+3x2^2, what are the gradient and Hessian, and how do their signs determine the extremum type?

Key Points

1
A critical point x0 for a local extremum must satisfy ∇f(x0)=0.
2
For twice continuously differentiable functions, a positive definite Hessian Hf(x0) guarantees an isolated local minimum.
3
A negative definite Hessian Hf(x0) guarantees an isolated local maximum.
4
For symmetric Hessians, definiteness can be checked by eigenvalues: all positive for positive definite, all negative for negative definite.
5
If the Hessian is indefinite (eigenvalues with mixed signs), the critical point is a saddle and there is no local extremum.
6
Quadratic examples become straightforward when the Hessian is diagonal, since eigenvalues equal the diagonal entries.
7
When Hessians are large, eigenvalue-based checks can be hard, motivating Sylvester’s Criterion as an alternative.

Highlights

Positive definite Hessians produce isolated local minima; negative definite Hessians produce isolated local maxima.

Mixed-sign eigenvalues make the Hessian indefinite, turning a critical point into a saddle with no local extremum.

For f(x1,x2)=x1^2+3x2^2, the Hessian [[2,0],[0,6]] is positive definite, so (0,0) is a (global) minimum.

For f(x1,x2)=x1^2−4x2^2, the Hessian [[2,0],[0,−8]] is indefinite, so (0,0) is a saddle point.

Eigenvalue checks are easy for diagonal symmetric Hessians but may be replaced by Sylvester’s Criterion for larger matrices.

Topics

Local Extrema
Hessian Test
Eigenvalues
Saddle Points
Sylvester’s Criterion