Minimum, maximum, and saddle point

Optimization: Extreme points

Minimum, maximum, and saddle point

The concepts of local minimum and local maximum are already known to us for functions of a single variable. The function $f(x)$ has a local minimum in $x=a$ if the graph near $x=a$ lies above $f(a)$ , more precisely, if there is an open interval $\ivoo{c}{d}$ around $a$ (that is to say, there are numbers $c\lt a$ and $d\gt a$ ) such that $f(x)\ge f(a)$ for all $x$ from $\ivoo{c}{d}$ . For the definition in the case of a function of two variables, we replace the open interval by an open disk.

Local extremes

Let $\epsilon$ be a positive number. The open disk around a point $p$ of ${\mathbb R}^2$ of radius $\epsilon$ is the subset $S_{p,\epsilon}$ of ${\mathbb R}^2$ consisting of all points $q\in {\mathbb R}^2$ at distance less than $\epsilon$ to $p$ . In a formula: $S_{p,\epsilon}=\left\{q\in{\mathbb R}^2\mid \sqrt{(p_1-q_1)^2 + (p_2-q_2)^2}\lt \epsilon\right\}$

Let $f$ be a bivariate function with domain $D$ and let $p$ be a point of $D$ .

The point $p$ is called a local minimum of $f$ if there is an open disk $S$ around $p$ (a set of the form $S=S_{p,\epsilon}$ ) for a suitable value of $\epsilon$ so $f(q)\ge f(p)$ for all $q\in D\cap S$ .
The point $p$ is called a local maximum of $f$ if there is an open disk $S$ around $p$ so $f(q)\le f(p)$ for all $q\in D\cap S$
The point $p$ is called a saddle point of $f$ if it is a stationary point, but in every open disk around $p$ there are points $q$ and $r$ such that $f(q)\gt f(p)$ and $f(r)\lt f(p)$ .

Points $p$ with $f(x)\le f(p)$ for all $x$ from the domain of $f$ are called maxima. Points $p$ with $f(x)\ge f(p)$ for all $x$ from the domain of $f$ are called minima.

Clearly, a maximum of $f$ will always be a local maximum and a minimum will always be a local minimum. In order to distinguish maxima and minima from local maxima and minima, we sometimes also call them global maxima and global minima.

Below you see the graph of the function $f(x,y)=\tfrac{1}{2}\!\left((1-(x-\tfrac{1}{2})^2-(y-\tfrac{1}{2})^2\right)$

maximum of f (x, y)

This function has the following partial derivatives: $f_x(x,y)=\tfrac{1}{2}-x\qquad\text{and}\qquad f_y(x,y)=\tfrac{1}{2}-y$ In the point $\rv{\tfrac{1}{2},\tfrac{1}{2}}$ , both derivatives are zero. This point is a stationary point. The function has a maximum value there.

The notion of saddle point is similar to the notion of inflection point for a stationary point of a function of a single variable.

For the distance between two points $p=\rv{p_1,p_2}$ and $q=\rv{q_1,q_2}$ in the plane we also write $||p-q||$ instead of $\sqrt{(p_1-q_1)^2 + (p_2-q_2)^2}$ . Using this notation the closed disk with center $p$ and radius $\epsilon$ can be defined as $\left\{q\in{\mathbb R}^2\mid||p-q||\le \epsilon\right\}$

Here is a generalization of the theorem Local extrema are stationary points for one variable.

If $f$ a bivariate differentiable function on a domain $D$ and $p$ is a local minimum or local maximum of $f$ , then $p$ is a stationary point of $f$ .

The proof is simple: Write $p=\rv{a,b}$ . Then $a$ is a local minimum or maximum of the differentiable function $f(a,y)$ in the variable $y$ . The theorem Local extrema are stationary points for the case of one variable gives $\left.\frac{\partial f(x,y)}{\partial y}\right|_{\rv{a,b}}=\left.\frac{\dd f(a,y)}{\dd y}\right|_{b}=0\tiny.$ Similarly, we have $\left.\frac{\partial f(x,y)}{\partial x}\right|_{\rv{a,b}}=\left.\frac{\dd f(x,b)}{\dd x}\right|_{a}=0\tiny.$ This means that $p=\rv{a,b}$ is a stationary point of $f$ .

Saddle points are by definition stationary points that are neither a local minimum nor a local maximum. The definition of a saddle point is even chosen so that a stationary point of $f$ is always a local minimum, local maximum, or saddle point.

ExampleAs with functions of one variable, for differentiable functions of two variables, the requirement that a point is a stationary point is necessary but not sufficient for a point to be a local minimum or a local maximum. We give a counterexample.

Below is the graph of the function $f(x,y)=\tfrac{1}{2}\!\left((1-(x-\tfrac{1}{2})^2+(y-\tfrac{1}{2})^2\right)$

saddle point

This function has the following partial derivatives: $f_x(x,y)=\tfrac{1}{2}-x\phantom{quad}\text{and}\phantom{quad}f_y(x,y)=y-\tfrac{1}{2}$ At the point $\rv{\tfrac{1}{2},\tfrac{1}{2}}$ both derivatives are zero again. But the coordinate line in the direction of the $x$ -axis is a downward parabola, and the coordinate line in the direction of the $y$ -axis is an upward parabola. Therefore, the point $\rv{\tfrac{1}{2},\tfrac{1}{2}}$ must be a saddle point.

The function $f(x,y)=-2\cdot x^2-2\cdot y\cdot x+36\cdot x-2\cdot y^2+42\cdot y-4$ has a local maximum. Which point is it?

$\rv{5, 8}$

Since a local maximum of a multivariate differentiable function is a stationary point, we first calculate the stationary points. The partial derivatives of $f$ are $f_x(x,y)=-4\cdot x-2\cdot y+36\phantom{quad}\text{and}\phantom{quad}f_y(x,y)=-2\cdot x-4\cdot y+42\tiny.$ The stationary points are the solutions of the system of equations $\lineqs{-4\cdot x-2\cdot y+36&=&0\cr -2\cdot x-4\cdot y+42&=&0\cr}$

This system has exactly one solution: ${x = 5\land y = 8}$ . We conclude that there is exactly one stationary point: $\rv{5, 8}$ . It is given that $f$ has a local maximum, so this point must be the answer: $\rv{5, 8}$ .

The graph of the function $f$ is shown in the figure below. The point $\rv{5,8,254}$ corresponding to the local maximum is indicated by a small black disk.

New example