# 4.7 Maxima/minima problems  (Page 4/10)

 Page 4 / 10

Use the second derivative to find the local extrema of the function

$f\left(x,y\right)={x}^{3}+2xy-6x-4{y}^{2}.$

$\left(\frac{4}{3},\frac{1}{3}\right)$ is a saddle point, $\left(-\frac{3}{2},-\frac{3}{8}\right)$ is a local maximum.

## Absolute maxima and minima

When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is bounded if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.

## Extreme value theorem

A continuous function $f\left(x,y\right)$ on a closed and bounded set $D$ in the plane attains an absolute maximum value at some point of $D$ and an absolute minimum value at some point of $D.$

Now that we know any continuous function $f$ defined on a closed, bounded set attains its extreme values, we need to know how to find them.

## Finding extreme values of a function of two variables

Assume $z=f\left(x,y\right)$ is a differentiable function of two variables defined on a closed, bounded set $D.$ Then $f$ will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:

1. The values of $f$ at the critical points of $f$ in $D.$
2. The values of $f$ on the boundary of $D.$

The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of $D,$ then it is located at an interior point of $D.$ But an interior point $\left({x}_{0},{y}_{0}\right)$ of $D$ that’s an absolute extremum is also a local extremum; hence, $\left({x}_{0},{y}_{0}\right)$ is a critical point of $f$ by Fermat’s theorem. Therefore the only possible values for the global extrema of $f$ on $D$ are the extreme values of $f$ on the interior or boundary of $D.$

## Problem-solving strategy: finding absolute maximum and minimum values

Let $z=f\left(x,y\right)$ be a continuous function of two variables defined on a closed, bounded set $D,$ and assume $f$ is differentiable on $D.$ To find the absolute maximum and minimum values of $f$ on $D,$ do the following:

1. Determine the critical points of $f$ in $D.$
2. Calculate $f$ at each of these critical points.
3. Determine the maximum and minimum values of $f$ on the boundary of its domain.
4. The maximum and minimum values of $f$ will occur at one of the values obtained in steps $2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}3.$

Finding the maximum and minimum values of $f$ on the boundary of $D$ can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in [link] . The same approach can be used for other shapes such as circles and ellipses.

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