Multivariable Extreme Values

What Counts

• $f$ has a local maximum at $P_0$ if $f(P)\le f(P_0)$ for all $|P_{0}P|<\delta$
• $f$ has a local minimum at $P_0$ if $f(P)\ge f(P_0)$ for all $|P_{0}P|<\delta$

Finding

• Critical points where $\vec{\nabla }f(P)=\vec{0}$ or DNE
• Categorization
  • ${\text{d}}^{2}f({\vec{r}}_0)>0,d\vec{r}\ne 0⟹\text{ local min at }{\vec{r}}_0$
  • ${\text{d}}^{2}f({\vec{r}}_0)<0,d\vec{r}\ne 0⟹\text{ local max at }{\vec{r}}_0$
  • ${\text{d}}^{2}f({\vec{r}}_0)=0,d\vec{r}\ne 0⟹\text{ not enough info}$
• Cases
  • Two Variable Extreme Values
  • More Than Two Variable Extreme Values

Multivariable |EVT

• Definition
  • $f$ is continuous on $D$
  • $D$ is bounded and closed
    • Bounded = can fit in a large enough finite ball → not infinite
    • Closed = Contains all its limit points → includes boundary
  • $⟹$ $f$ takes it’s extreme values on $D$
    • $\exists P_1,P_2\in D\text{ s.t. }\forall P\in Df(P_1)\le f(P)\le f(P_2)$
• Steps
  1. Find critical points in $D⟺\vec{\nabla }f=\vec{0}⟹P_1,P_2,\dots$
  2. Find max/min values on the boundary of $D$ (${\min }_{\partial D}f,{\max }_{\partial D}f$)
     • $\partial D$ means the boundary of $D$ for some reason
     • Can either solve by parameterizing or using Lagrange multipliers
  3. Compare the two
     • ${\max }_{D}f=\max \{f(P_1),\dots ,f(P_n),{\max }_{\partial D}f\}$
     • ${\min }_{D}f=\min \{f(P_1),\dots ,f(P_n),{\min }_{\partial D}f\}$

Lagrange Multiplier Method

Alternate method to find critical points

Requirements

Where $f$ is the function and $g=c$ is the constraint

1. $f$ and $g$ have continuous partial derivatives
2. $\vec{\nabla }g\ne 0$ on $g(x,y)=0$
3. $f$ has a local extremum at $P_0$ on $g(x,y)=0$

Usage

Solve the following system for $P_0$ to get the critical point

$$\begin{gathered} \vec{\nabla }f(P_0)=\lambda \vec{\nabla }g(P_0) \\ g(P_0)=c \end{gathered}$$

where $\lambda$ is a constant

Two or More Constraints

$$\begin{array}{rl}\vec{\nabla }f(P_0)& ={\lambda }_1\vec{\nabla }{g}_1(P_0)+{\lambda }_2\vec{\nabla }{g}_2(P_0)+\dots \\ g_1(P_0)& =c_1\\ g_2(P_0)& =c_2\\ ⋮& \end{array}$$