4.7 Maxima/minima problems  (Page 4/10)

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.

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.

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 $D.$

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.