Stationary points

Optimization: Extreme points

Stationary points

The concepts of stationary point, minimum, and maximum are already known 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 are numbers $c\lt a$ and $d\gt a$ such that $f(x)\ge f(a)$ for all $x$ from $\ivoo{c}{d}$ . For a differentiable function $f(x)$ , a local minimum (or maximum) is a stationary point, that is, a point $x=a$ at which the tangent line to $f$ is horizontal, in other words $f'(a)=0$ .

As with functions of one variable we will examine local minima and maxima of bivariate functions. We start with the two-dimensional counterpart of the concept of stationary point.

Stationary point

Let $f$ be a bivariate differentiable function. A point $\rv{a,b}$ is a stationary point of the function $f(x,y)$ if all partial derivatives of $f$ at this point are equal to zero.

Stationary points can be found by solving the following system of equations: $\eqs{f_x(x,y)&=&0\cr f_y(x,y)&=&0\cr}$

Later we will see that, if $\rv{a,b}$ is a stationary point of $f$ , the tangent plane to the graph of $f$ at $\rv{a,b,f(a,b)}$ is horizontal.

Calculate the stationary points of the function $f(x,y)=-5\cdot y\cdot x^2+y^2\cdot x+3\cdot y\cdot x\tiny.$

Enter your answer as a set of row vectors: $\left\{\rv{a,b},\rv{c,d},\ldots\right\}$ for appropriate numbers $a$ , $b$ , $c$ , $d$ , $\ldots$

$\left\{\rv{0,0},\rv{{{3}\over{5}},0},\rv{0,-3}, \rv{{{1}\over{5}},-1}\right\}$

The partial derivatives of $f$ are $f_x(x,y)=-10\cdot y\cdot x+y^2+3\cdot y\phantom{quad}\text{and}\phantom{quad}f_y(x,y)=-5\cdot x^2+2\cdot y\cdot x+3\cdot x\tiny.$ The stationary points are the solutions of the system of equations $\lineqs{-10\cdot y\cdot x+y^2+3\cdot y&=&0\cr -5\cdot x^2+2\cdot y\cdot x+3\cdot x&=&0\cr}$ To solve this system, we split it into smaller systems by factoring the left members of the equations. We find: $\lineqs{y\cdot \left(-10\cdot x+y+3\right)&=&0\cr x\cdot \left(-5\cdot x+2\cdot y+3\right)&=&0\cr}$ Apparently, each of the equations can be split into two simpler equations. Combination of the four possibilities leads to four solutions:

$\eqs{x&=&0\cr y&=&0\cr}\phantom{xx} \implies$ $\phantom{xx} \rv{x,y}=\rv{0,0}$
$\eqs{ -5\cdot x+2\cdot y+3&=&0\cr y&=&0\cr} \phantom{xx} \implies$ $\phantom{xx} \rv{x,y}=\rv{{{3}\over{5}},0}$
$\eqs{ x&=&0\cr -10\cdot x+y+3&=&0\cr}\phantom{xx} \implies$ $\phantom{xx} \rv{x,y}=\rv{0,-3}$
$\eqs{ -10\cdot x+y+3&=&0\cr -5\cdot x+2\cdot y+3&=&0\cr}\phantom{xx}\implies$ $\phantom{xx} \rv{x,y}=\rv{{{1}\over{5}},-1}$

We conclude that there are four stationary points: $\left\{\rv{0,0},\rv{{{3}\over{5}},0},\rv{0,-3}, \rv{{{1}\over{5}},-1}\right\}$ .

The graph of the function $f$ is shown in the figure below. The points of the graph associated with $\left\{\rv{0,0},\rv{{{3}\over{5}},0},\rv{0,-3}, \rv{{{1}\over{5}},-1}\right\}$ are marked with a small black disk.

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