A function $f(x_1, x_2, \ldots, x_n)$ has a local maximum or minimum at a point $(x_1^*, x_2^*, \ldots, x_n^*)$ if this point is one of the following:

• $(x_1^*, x_2^*, \ldots, x_n^*)$ is a critical point: $\nabla f(x_1^*, x_2^*, \ldots, x_n^*) = 0$
• $(x_1^*, x_2^*, \ldots, x_n^*)$ is a singular point: $\nabla f(x_1^*, x_2^*, \ldots, x_n^*)$ does not exist
• $(x_1^*, x_2^*, \ldots, x_n^*)$ is a boundary point: $(x_1^*, x_2^*, \ldots, x_n^*)$ is on the boundary of the domain of $f$

If a function $f(x_1, x_2, \ldots, x_n)$ is continuous on a closed, bounded set $D \subset \mathbb{R}^n$, then $f$ attains both a global maximum and a global minimum on $D$.

Let $(x_1^*, x_2^*, \ldots, x_n^*)$ be a critical point of a function $f(x_1, x_2, \ldots, x_n)$. If $f$ is twice differentiable in a neighborhood of $(x_1^*, x_2^*, \ldots, x_n^*)$, then:

• If all eigenvalues of $H_f(x_1^*, x_2^*, \ldots, x_n^*)$ are strictly positive, then $f$ has a local minimum at $(x_1^*, x_2^*, \ldots, x_n^*)$
• If all eigenvalues of $H_f(x_1^*, x_2^*, \ldots, x_n^*)$ are strictly negative, then $f$ has a local maximum at $(x_1^*, x_2^*, \ldots, x_n^*)$
• If $H_f(x_1^*, x_2^*, \ldots, x_n^*)$ has both positive and negative eigenvalues, then $f$ has a saddle point at $(x_1^*, x_2^*, \ldots, x_n^*)$
• If $H_f(x_1^*, x_2^*, \ldots, x_n^*)$ has at least one eigenvalue equal to zero, and the other eigenvalues have the same sign, then the test is inconclusive

Let $(x_1^*, x_2^*)$ be a critical point of a function $f(x, y)$. If $f$ is twice differentiable in a neighborhood of $(x_1^*, x_2^*)$, then:

• If $D(x_1^*, x_2^*) > 0$ and $f_{xx}(x_1^*, x_2^*) > 0$, then $f$ has a local minimum at $(x_1^*, x_2^*)$
• If $D(x_1^*, x_2^*) > 0$ and $f_{xx}(x_1^*, x_2^*) < 0$, then $f$ has a local maximum at $(x_1^*, x_2^*)$
• If $D(x_1^*, x_2^*) < 0$, then $f$ has a saddle point at $(x_1^*, x_2^*)$
• If $D(x_1^*, x_2^*) = 0$, then the test is inconclusive

Let $f(x, y)$ and $g(x, y)$ be differentiable functions. If $f$ has a local extremum at $(x^*, y^*)$ subject to the constraint $g(x, y) = c$, and if $\nabla g(x^*, y^*) \neq (0, 0)$, then there exists a scalar $\lambda$ such that $\nabla f(x^*, y^*) = \lambda \nabla g(x^*, y^*)$.