Lecture 43: Local and Absolute Extrema of Multivariable Functions

Section Overview

(1) Unconstrained extremum
(2) Conditional Extrema and Lagrange Multiplier Method
(3) Maximum and minimum values

Frequently Tested Question Types in This Section

Problem Type 1: Find Extrema (Unconditional)
Problem Type 2: Find the maximum and minimum values
Question Type 3: Word problems with maximum and minimum values

43.1 Unconditional Extrema of Multivariable Functions

43.1.1 Basic Concepts

#####Definition: Extremumofmultivariablefunctions

Description: the maximum or minimum value of a multivariable function in a neighborhood;

If in a neighborhood of point $(x_0,y_0)$, the inequality always holds: $$f(x,y)\leq f(x_0,y_0)\quad(f(x,y)\geq f(x_0,y_0))$$

Then $f$ has a local maximum (minimum) at $(x_0,y_0)$, and $(x_0,y_0)$ is called a local maximum (minimum) point.

The maximum and minimum are collectively called extremums, and the maximum and minimum points are collectively called extremum points;

Explanation

Give an example
$z=3x^{2}+4y^{2}$ is minimal at $(0.0)$;
$z=-\sqrt{x^{2}+y^{2}}$ is maximum at $\quad(0.0)$

#####Definition: Thestationarypointofamultivariablefunction

Description: point where the partial derivative of a multivariable function equals 0 is called the stationary point of the multivariable function;

Explanation

$f_{x}^{\prime}=0.f_{y}^{\prime}=0$ Points that exist simultaneously are called stationary points;

#####Theorem: Thenecessaryconditionfortheexistenceofextremainmultivariablefunctions

Description: if $z=f(x,y)$ has a partial derivative at point $(x_0,y_0)$ and takes an extremum at $(x_0,y_0)$, then we have: $$f_{x}^{\prime}(x_{0},y_{0})=0,\,f_{y}^{\prime}(x_{0},y_{0})=0$$

Explanation

Requirements:
The two first-order partial derivatives of a point must be zero; this is a necessary condition for obtaining an extremum, butnot a sufficient condition;
Narrow the scope;
The relationship between stationary points and extrema points:
Generally:
Concept:
Not all stationary points are necessarily extremal; Extremal points are not necessarily stationary points <- For example, $|x|+|y|$;
Conclusion:
Two scenarios: possible extreme points
(1) Stationary <- Mainly this situation;
(2) Neither $f_x$ nor $f_y$ exists
(3)$f_x$ Nonexistent, $f_y=0$
(4)$f_y$ Nonexistent, $f_x=0$
Premises of differentiability:
Concept:
If $z=f(x,y)$ is differentiable, the extremum must be a stationary point;
Conclusion:
Extremes only need to find stationary points;

#####Theorem: Sufficientconditionsforextremes

Description: Determination of Standpoint Extremity of Multivariable Functions

When $z=f(x,y)$ has a second-order continuous partial derivative in a neighborhood of point $P_0(x_0,y_0)$, and $f_x^{\prime}(x_0,y_{0})=f_y^{\prime}(x_0,y_{0})=0$, $A=f_{xx}^{\prime\prime}(x_{0},y_{0})\quad\quad B=f_{xy}^{\prime\prime}(x_{0},y_{0})\quad\quad C=f_{yy}^{\prime\prime}(x_{0},y_{0})$;

1. If it is $AC-B^{2}>0$, then there are extremes

1. A<0 is the maximum;

2. A>0 is the minimum value;

2. If it is $AC-B^{2}<0$, then there is no extreme value

3. If it is $AC-B^{2}=0$, then it can only be determined by definition

Explanation: Summary

After finding the stationary point, you only need to use this sufficient conditional determination method to determine whether it is an extremum

43.1.2 Example Problems

Example: Find the extremum of $f(x,y)=x^{3}-y^{3}+3x^{2}+3y^{2}-9x$;

Analysis
Analysis
Finding the partial derivative separately:
$f_{x}^{\prime}=3x^{2}+6x-9$
$f_{y}^{\prime}=-3y^{2}+6y$
Gain
$\begin{cases}x^{2}+2x-3=0\\-y^{2}+2y=0\end{cases}$
Simplify
$(x+3)(x-1)=0$ hour, $x=-3\text{ or }1$;
$y(y-2)=0$ hour, $y=0\text{ or }2$;
Permutations and combinations to obtain stationary points
$(1,0),(1,2),(-3,0),(-3,2)$
Find A, B, C
$f_{xx}^{\prime\prime}=6x+6$
$f_{xy}^{\prime\prime}=0$
$f_{yy}^{\prime\prime}=-6y+6$
When (1,0) point:
$A=12.B=0.C=6$
So $AC-B^{2}=72>0$
Therefore, it is the minimum value, that is, after substituting $(1,0)$, the result is -5, which is the minimum value;
Other analogies
Question Type: Extremumofmultivariablefunctions

43.2 Conditional Extremum of Multivariable Functions: Lagrange Multiplier Method

43.2.1 Basic Concepts

Unconditional Extrema and Conditional Extremum

Unconditional
Example: Find the tallest person in the entire school
Conditional
Example: Find the tallest student in the school, from Shandong, Gemini;
The more conditions you add, the smaller the maximum value

#####Definition: Conditionextremes

Description: given $z=f(x,y)$, its extremum is under the condition of $\varphi(x,y)=0$, find its extremum as the conditional extremum;

Explanation

Difference from unconstrained extremums:
Unconstrained extremum: an extreme value within the domain, i.e., the highest or lowest point within the domain or locally;
Constrained extrema: The content of constraints $\varphi (x, y)=0$ is usually a line geometrically, i.e., on a curve, local minimum and local maximum;
Example:
In $z=f(x,y)$: $z=f(x,\varphi(x))$
Here, derivative from z to x: $\frac{dz}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}$
Write out $\frac{dz}{dx}\vert_{x=x_{0}}=f^{\prime}(x_{0},y)+f^{\prime}(x_{0},y)\frac{dy}{dx}\vert_{x=x_{0}}=0$

#####Theorem: Lagrangemultipliermethod

description:

(1) For a single constraint: let $F(x,y,\lambda)=f(x,y)+\lambda\varphi(x,y)$

then the necessary condition to obtain the extremum is: $$\begin{cases}F_x=f_x^{\prime}(x,y)+\lambda\varphi_x^{\prime}(x,y)=0,\\F_y=f_y^{\prime}(x,y)+\lambda\varphi_y^{\prime}(x,y)=0,\\F_\lambda=\varphi(x,y)=0,\end{cases}$$

(2) When multiple constraints are present: the extreme value of the condition under $f(x,y,z)\text{ In terms of conditions }\varphi(x,y,z)=0,\psi(x,y,z)=0$ conditions

Order $F(x,y,z,\lambda,\mu)=f(x,y,z)+\lambda\varphi(x,y,z)+\mu\psi(x,y,z)$

Explanation

Explanation:
$\lambda$ is a parameter
Meaning:
Transform the necessary condition for obtaining conditional extrema of a binary function under constraints into a large F Problem of the necessary condition for unconditional extrema of such a ternary function -> For three partial derivatives, equal to 0;
Note:
The point obtained by solving this constraint is the possible extremum, not a definite one;

#####Theorem: nMeta-Lagrange multiplier method

Description: the above methods can be generalized to the extremum problem of $n$-element functions under $m$ constraints,

To find the extremum of $u=f(x,y,z)$ under condition $\varphi(x,y,z)=0,\psi(x,y,z)=0$, construct the Lagrange function: $$F(x,y,z,\lambda,\mu)=f(x,y,z)+\lambda\varphi(x,y,z)+\mu\psi(x,y,z).$$ find the partial derivatives for $x,y,z,\lambda,\mu$ of $F$ and construct a system of equations: $$\begin{aligned}&f_{x}^{\prime}(x,y,z)+\lambda\varphi_{x}^{\prime}(x,y,z)+\mu\psi_{x}^{\prime}(x,y,z)=0,\\&f_{y}^{\prime}(x,y,z)+\lambda\varphi_{y}^{\prime}(x,y,z)+\mu\psi_{y}^{\prime}(x,y,z)=0,\\&f_{z}^{\prime}(x,y,z)+\lambda\varphi_{z}^{\prime}(x,y,z)+\mu\psi_{z}^{\prime}(x,y,z)=0,\\&\varphi(x,y,z)=0,\\&\psi(x,y,z)=0.\end{aligned}$$ solve for $x,y,z,\lambda$ and $\mu$, then ($x,y,z)$ is thepossible extreme point.

43.3 Maximum/Minimum Values of Multivariable Functions

43.3.1 Basic Concepts

Summary: Find the maximum and minimum values of the continuous function f(x, y) on a bounded closed domain D

Step 1: Find the possible extremum points of f(x, y) within D;
(1) Stationing point;
(2) Two first-order partial derivatives, at least one point that does not exist
Step 2: Find the maximum and minimum values of f(x, y) on theboundaryof D;
Find the maximum and minimum -> on the boundary. This is essentially finding a conditional extremum;
Step 3: Compare the extremes;

44.3.2 Conditional Maximum Problem

43.3 Frequently Tested Question Types

Question Type: Findtheextremum

PART 1: Problem-solving methods

Problem Type: Know the total differential of a function and determine the extremum

Solution Guidance:
Usually, Method Three or Method One is used first;
Method 1: Sufficient conditions for extremes
Use sufficient conditions for extreme values. First, use the full differential to know the partial derivatives for x and y, then use the partial derivatives to find the values of A、B、C and judge using formulas;
Method 2: Partial integration
Core: Know the total differential -> Use partial integrals to find the function;
Example: When the total derivative of a binary function $z=f(x,y)$ is $dz = xdx + ydy$, its partial derivative can be found
$z_{x}=x\quad\quad z=\int xdx=\frac{1}{2}x^{2}+q(y)$
$z_{y}=y \quad\quad z_{y}=q^{\prime}(y)=y,q(y)=\frac{1}{2}y^{2}+c$
So $z=\frac{1}{2}(x^{2}+y^{2})+c$
Method 3: Derivative Derivation
Core: Given a total differential, reconstruct its potential function.
Example: $dz=xdx+ydy=d(\frac{1}{2}x^2)+d(\frac{1}{2}y^{2})=d(\frac{1}{2}(x^{2}+y^{2}))$
Since the differentials are equal, the two can only differ by constant, so $z=\frac{1}{2}(x^{2}+y^{2})+c$

Problem Type: Know the form of the function and find the extrema point

(1.1) No possible extreme points provided:
Compute the partial derivatives of the function x and y respectively so that the partial derivative equals 0, and find possible stationary points (if no possible extrema points are given)
(1.2) Given possible extreme points:
Use necessary conditions to determine -> determine whether it is a stationary point:
Necessary conditions for the existence of extreme points -> $f_{x}^{\prime}(x_{0},y_{0})=0,\,f_{y}^{\prime}(x_{0},y_{0})=0$
(2) Sufficient condition judgment: Use the ABC formula to judge
Note:
If calculating ABC is troublesome, you can use "substitution first, then calculate"-> When finding Y, first substitute X; similarly, when calculating X;

PART 2: Typical Example Problems

PART 3: Key Points Review

Question Type: ConditionalextremumandLagrangemultipliermethod

PART 1: Problem-solving methods

Question Type: Conditional maximum-value problems

Concept:
Using the Lagrange multiplier method, first find all possible extrema points of the function, then compare all possible extrema points without determination to obtain the maximum;
Steps to solve the problem:
(1) Construct the Lagrange function;
(2) Find partial derivatives;
(3) Find a station;
Obtain several equations equal to 0, and use them to solve for possible values of x and y;
(4) Compare each point to determine the maximum and minimum values;

Problem Type: Given the full differential, find the maximum

Method:
(1) First, recover the potential function from the total differential.
Two methods -> partial integral, derivative;
(2) Find the minimum point of the region;
1. Possible extreme points within the region;
2. The maximum and minimum values on the boundary curve;
Method One:
Using conditions, convert extremes into unconstrained extremums;
Method Two:
The general method is the Lagrange multiplier method;
Method 3:
Parametric equations
3. Compare the maximum and minimum values on the boundary with possible extremes within the region;