Extrema

Local

Let $f(x,y)$ be defined on a region containing the point $(a,b)$

• $f(a,b)$ is a local max of $f$ if $f(a,b)\ge f(x,y)$ for all domain points $(x,y)$ in a disk centered at $(a,b)$
• $f(a,b)$ is a local min of $f$ if $f(a,b)<f(x,y)$ for all domain points $(x,y)$ in a disk centered at $(a,b)$

Global

Let $f(x,y)$ be defined on a region containing the point $(a,b)$

• $f(a,b)$ is a global max of $f$ if $f(a,b)\ge f(x,y)$ for all $(x,y)$ in the domain of $f$
• $f(a,b)$ is a global min of $f$ if $f(a,b)<f(x,y)$ for all $(x,y)$ in the domain of $f$

Theorem

On a Closed and Bounded Domain, any continous function $f$ attains a global minimum and maximum

Strategy for finding global extrema

Let $f=f(x,y)$, domain $R$ be closed and bounded.

1. Find all critical points of $f$ inside $R$, and on the Boundary of $R$
2. Evaluate $f$ at each critical point, as well as any endpoints of the boundary
3. The smallest value found is the global minimum; the largest is the global maximum

Critical Point

An extremum or saddle point. Given $f(x,y)$, a point $(a,b)$ in the domain of $f$ is called a Critical Point when

$$\nabla f(a,b)=\left[\begin{array}{c}0\\ 0\end{array}\right]⟺Df(a,b)=[0,0]$$ $$\nabla f(a,b)=\text{DNE}⟺Df(a,b)=\text{DNE}$$

Second Derivative Test

This is for classification of critical points Suppose $(a,b)$ is a critical point of $f(x,y)$ and $f$ has continuous second partial derivatives. Then we have:

• If $\det (H_f(a,b))>0$ and $f_{xx}(a,b)>0$, then $f(a,b)$ is a local minimum.
• If $\det (H_f(a,b))>0$ and $f_{xx}(a,b)<0$, then $f(a,b)$ is a local maximum.
• If $\det (H_f(a,b))<0$, then $f$ has a saddle point at $(a,b)$.
• If $\det (H_f(a,b))=0$, the test is inconclusive.

$$\det (H_f(a,b))=f_{xx}{f}_{yy}-f_{xy}^2{|}_{(a,b)}$$

General Test

More generally, if $f:{\mathbb{R}}^n\to \mathbb{R}$ has a critical point at $P$ then:

• If all eigenvalues of $H_f(P)$ are positive, $f$ is concave up in every direction from $P$ and so has a local minimum at $P$.
• If all eigenvalues of $H_f(P)$ are negative, $f$ is concave down in every direction from $P$ and so has a local maximum at $P$.
• If some eigenvalues of $H_f(P)$ are positive and some are negative, $f$ is concave up in some directions from $P$ and concave down in others, so has neither a local minimum nor maximum at $P$.
• If all eigenvalues of $H_f(P)$ are positive or zero, $f$ may have either a local minimum or neither at $P$.
• If all eigenvalues of $H_f(P)$ are negative or zero, $f$ may have either a local maximum or neither at $P$.

Constraints of Equality

The Objective Function (the function containing the extrema) can be constrained by constraint function(s)

• For some simple problems you can just use substitution.
• For for complicated problems, the method of Lagrange Multipliers can be used to convert it into an unconstrained problem whose number of variables is the original number of variables plus the original number of equality constraints.

  Lagrange Multipliers

  Extrema of $f(\vec{r})$ subject to constraint $g(\vec{r})=c$ satisfy $(\nabla f=\lambda \nabla g\wedge g=c)$ Extrema of $f(\vec{r})$ subject to constraints $g(\vec{r})=c_1,h(\vec{r})=c_2$ satisfy $(\nabla f=\lambda \nabla g+\mu \nabla h\wedge g=c_1,h=c_2)$

  Examples

  Example 1

  Find the points on surface $z^2=xy+4$ that are closest to the origin. Let $d(x,y,z)=\sqrt{x^2+y^2+z^2}$ be the distance from a point to the origin Let $f(x,y,z)=x^2+y^2+z^2$ be the proxy to the distance function that we will use to solve this problem, because its extrema exactly correspond to that of $d$. $f$ is our objective function Let $g(x,y,z)=z^2-xy=4$ be our constraint function Goal: find points where $\nabla f\propto \nabla g$ and $g=c$ Find Gradients: $\nabla f=<2x,2y,2z>\nabla g=<-y,-x,2z>$ Solve this system:

  1. $2x=-\lambda y$
  2. $2y=-\lambda x$
  3. $2z=\lambda 2z$
  4. $z^2-xy=4$ Blah blah blah… You do a bunch of different cases and get the following points: $(2,-2,0),(-2,2,0),(0,0,2),(0,0,-2)$ You can plug each of these into $f$ and you’ll find $\max f=8⟹\max d=\sqrt{2}$ and $\min f=4⟹\min d=2$

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