# 4.7 Maxima/minima problems  (Page 2/10)

 Page 2 / 10

Find the critical point of the function $f\left(x,y\right)=x3+2xy-2x-4y.$

$\left(2,-5\right)$

The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.

## Definition

Let $z=f\left(x,y\right)$ be a function of two variables that is defined and continuous on an open set containing the point $\left({x}_{0},{y}_{0}\right).$ Then f has a local maximum at $\left({x}_{0},{y}_{0}\right)$ if

$f\left({x}_{0},{y}_{0}\right)\ge f\left(x,y\right)$

for all points $\left(x,y\right)$ within some disk centered at $\left({x}_{0},{y}_{0}\right).$ The number $f\left({x}_{0},{y}_{0}\right)$ is called a local maximum value . If the preceding inequality holds for every point $\left(x,y\right)$ in the domain of $f,$ then $f$ has a global maximum (also called an absolute maximum ) at $\left({x}_{0},{y}_{0}\right).$

The function $f$ has a local minimum at $\left({x}_{0},{y}_{0}\right)$ if

$f\left({x}_{0},{y}_{0}\right)\le f\left(x,y\right)$

for all points $\left(x,y\right)$ within some disk centered at $\left({x}_{0},{y}_{0}\right).$ The number $f\left({x}_{0},{y}_{0}\right)$ is called a local minimum value . If the preceding inequality holds for every point $\left(x,y\right)$ in the domain of $f,$ then $f$ has a global minimum (also called an absolute minimum ) at $\left({x}_{0},{y}_{0}\right).$

If $f\left({x}_{0},{y}_{0}\right)$ is either a local maximum or local minimum value, then it is called a local extremum (see the following figure).

In Maxima and Minima , we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

## Fermat’s theorem for functions of two variables

Let $z=f\left(x,y\right)$ be a function of two variables that is defined and continuous on an open set containing the point $\left({x}_{0},{y}_{0}\right).$ Suppose ${f}_{x}$ and ${f}_{y}$ each exists at $\left({x}_{0},{y}_{0}\right).$ If $f$ has a local extremum at $\left({x}_{0},{y}_{0}\right),$ then $\left({x}_{0},{y}_{0}\right)$ is a critical point of $f.$

## Second derivative test

Consider the function $f\left(x\right)={x}^{3}.$ This function has a critical point at $x=0,$ since $f\prime \left(0\right)=3{\left(0\right)}^{2}=0.$ However, $f$ does not have an extreme value at $x=0.$ Therefore, the existence of a critical value at $x={x}_{0}$ does not guarantee a local extremum at $x={x}_{0}.$ The same is true for a function of two or more variables. One way this can happen is at a saddle point    . An example of a saddle point appears in the following figure.

In this graph, the origin is a saddle point. This is because the first partial derivatives of $f\left(x,y\right)={x}^{2}-{y}^{2}$ are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to $y=0$ is $z={x}^{2}$ (a parabola opening upward), but the vertical trace corresponding to $x=0$ is $z=\text{−}{y}^{2}$ (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.

## Definition

Given the function $z=f\left(x,y\right),$ the point $\left({x}_{0},{y}_{0},f\left({x}_{0},{y}_{0}\right)\right)$ is a saddle point if both ${f}_{0}\left({x}_{0},{y}_{0}\right)=0$ and ${f}_{y}\left({x}_{0},{y}_{0}\right)=0,$ but $f$ does not have a local extremum at $\left({x}_{0},{y}_{0}\right).$

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