4.5 KiB
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), thenf
has a local minimum at\mathbf{x^*}
. - If
H_f(\mathbf{x^*})
is negative definite (all eigenvalues are negative), thenf
has a local maximum at\mathbf{x^*}
. - If
H_f(\mathbf{x^*})
is indefinite (both positive and negative eigenvalues), thenf
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}).