mathematics-physics-wiki/docs/en/mathematics/multivariable-calculus/extrema.md

Extrema

Definition: for D \subseteq \mathbb{R}^n let f: D \to \mathbb{R} be differentiable and D contains no boundary points (open). A point \mathbf{x^*} \in D is called a critical point for f \iff \nabla f(\mathbf{x^*}) = \mathbf{0}.

Definition: f has (strict) global \begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix} in \mathbf{x^*} \in D \iff \forall \mathbf{x} \in D \backslash \{\mathbf{x^*}\} \Big[f(\mathbf{x^*}) \begin{matrix} (>) \\ \geq \\ \leq \\ (<) \end{matrix} f(\mathbf{x}) \Big].

Definition: f has (strict) local \begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix} in \mathbf{x^*} \in D \iff \exists r_{>0} \forall \mathbf{x} \in D \backslash \{\mathbf{x^*}\} \Big[f(\mathbf{x^*}) \begin{matrix} (>) \\ \geq \\ \leq \\ (<) \end{matrix} f(\mathbf{x}) \;\land\; (0) < \|\mathbf{x} - \mathbf{x^*}\| < r \Big]

Theorem: if f has local \begin{matrix} \text{ maximum } \\ \text{ minimum } \end{matrix} at \mathbf{x^*} \in D then \mathbf{x^*} is a critical point of for f.

??? note "Proof:"

Will be added later.

A second derivative test

Definition: suppose f: \mathbb{R}^n \to \mathbb{R} is differentiable with \mathbf{x} \in \mathbb{R}^n. The Hessian matrix of f is defined as

H_f(\mathbf{x}) := \begin{pmatrix} \partial_{11} f(\mathbf{x}) & \dots & \partial_{1n} f(\mathbf{x}) \ \vdots & \ddots & \vdots \ \partial_{n1} f(\mathbf{x}) & \dots & \partial_{nn} f(\mathbf{x}) \end{pmatrix}.

Theorem:

• If H_f(\mathbf{x^*}) is positive definite (all eigenvalues are positive), then f has a local minimum at \mathbf{x^*}.
• If H_f(\mathbf{x^*}) is negative definite (all eigenvalues are negative), then f has a local maximum at \mathbf{x^*}.
• If H_f(\mathbf{x^*}) is indefinite (both positive and negative eigenvalues), then f has a saddle point at \mathbf{x^*}.
• If H_f(\mathbf{x^*}) is neither positive nor negative definite, nor indefinite, (eigenvalues equal to zero) this test gives no information.

??? note "Proof:"

Will be added later.

Extrema on restricted domains

Theorem: let D \subseteq \mathbb{R}^n be bounded and closed (D contains all boundary points). Let f: D \to \mathbb{R} be continuous, then f has a global maximum and minimum.

??? note "Proof:"

Will be added later.

Procedure to find the global maximum and minimum:

• Find critical points in the interior.
• Find global extrema on the boundary.
• Find the largest/smallest among them.

Lagrange multipliers

Theorem: let f: M \to \mathbb{R} and g: \mathbb{R}^n \to \mathbb{R} with M the boundary of D given by

M := \big{\mathbf{x} \in \mathbb{R}^n ;\big|; g(\mathbf{x}) = 0 \big} \subseteq D,

suppose that there is global maximum or minimum \mathbf{x^*} \in M of f that is not an endpoint of M and \nabla g(\mathbf{x^*}) \neq \mathbf{0}. Then there exists a \lambda^* \in \mathbb{R} such that (\mathbf{x^*}, \lambda^*) is a critical point of the Lagrange function

L(\mathbf{x}, \lambda) := f(\mathbf{x}) - \lambda g(\mathbf{x}).

??? note "Proof:"

Will be added later.

The general case

Theorem: Let f: S \to \mathbb{R} and \mathbf{g}: \mathbb{R}^m \to \mathbb{R}^n with m \leq n -1 restrictions given by

S := \big{\mathbf{x} \in \mathbb{R}^n ;\big|; \mathbf{g}(\mathbf{x}) = 0 \big} \subseteq D,

suppose that there is global maximum or minimum \mathbf{x^*} \in S of f that is not an endpoint of S and D \mathbf{g}(\mathbf{x^*}) \neq \mathbf{0}. Then there exists a \mathbf{\lambda^*} \in \mathbb{R^m} such that (\mathbf{x^*}, \mathbf{\lambda^*}) is a critical point of the Lagrange function

L(\mathbf{x}, \mathbf{\lambda}) := f(\mathbf{x}) - \big\langle \mathbf{\lambda},; \mathbf{g}(\mathbf{x}) \big\rangle.

??? note "Proof:"

Will be added later.

Example

Let f: M_1 \cap M_2 \to \mathbb{R} and g_{1,2}: \mathbb{R}^n \to \mathbb{R} with the restrictions given by

M_{1,2} := \big{\mathbf{x} \in \mathbb{R}^n ;\big|; g_{1,2}(\mathbf{x}) = 0 \big} \subseteq D,

suppose that there is global maximum or minimum \mathbf{x^*} \in M_1 \cap M_2 of f that is not an endpoint of M_1 \cap M_2 and \nabla g_{1,2}(\mathbf{x^*}) \neq \mathbf{0}. Then there exists a \lambda_{1,2}^* \in \mathbb{R} such that (\mathbf{x^*}, \lambda_{1,2}^*) is a critical point of the Lagrange function

L(\mathbf{x}, \lambda_1, \lambda_2) := f(\mathbf{x}) - \lambda_1 g_1(\mathbf{x}) - \lambda_2 g_2(\mathbf{x}).