Optimization Now that we have discussed local maximums and minimums of functions, it is natural to consider applying these concepts to actual problems. In addition to just finding a local maximum or minimum, we might want to find the largest or smallest value on a region R. These values are called the global maximum and global minimum of the function on the region R. That is, we have Global Maximums and Global Minimums of Functions f(x, y) has a global maximum on a region R at the point (x0, y0) provided f(x0, y0) ¥ f(x, y) for all points (x, y) in R f(x, y) has a global minimum on a region R at the point (x0, y0) provided f(x0, y0) § f(x, y) for all points (x, y) in R A natural question is whether or not a function has a global maximum or global minimum value on a region R. Thinking back to one-variable calculus, the function y = x2 has a global minimum at (0, 0), but goes off to infinity as the x-values get further away from 0. But if we specify a set of x-values, such as [-1, 2], then we can say that the function has a global maximum on [-1, 2]. In particular, the maximum occurs at x = 2. There was a result in one-variable calculus that showed that a continuous function (of one-variable) has a global maximum and a global minimum on [a, b] for constants a and b. An analogous result holds for functions of two or more variables. Extreme Value Theorem for Multivariable Functions If f(x, y) is a continuous function on a closed and bounded region R, then f(x, y) has a global maximum at some point (x1, y1) in R and a global minimum at some points (x2, y2) in R. A region R is said to be closed if the region contains is boundary. (In terms of a graph of the region, the region is outlined with solid lines as compared to dashed lines. This is analogous to requiring an interval [a, b] in one-variable calculus as opposed to (a, b), which did not include the end points.) A region is bounded if the values of the functions are finite (not infinity). 1Notice that the theorem does not tell us the locations of the points (x1, y1) and (x2, y2), but rather only that we can find them. The next natural question to ask is how we find those points. There are a lot (an infinite number) to choose from potentially. However, we only need to find critical points inside the region R and look for maximums and minimums (usually, but not always, critical points) on the boundary of R. The global maximum and the global minimum will occur from this collection of points. This should sound familiar to how one found global maximums and global minimums in one-variable calculus. As a reminder, here is the procedure we used before. How To Find Global Maximum and Global Minimum of a One-Variable Function on an interval [a, b] 1. Take the derivative of the function, set it equal to 0 and solve for x. These are critical points of the function. 2. Classify the critical points that are in [a, b] found in (1) as a local max/local min or neither. (We ignore any other critical points.) 3. Find the y-values of the critical points in (2), as well as the y-values of the endpoints. 4. The largest y-value(s) corresponds to the global maximum and the smallest y-value(s) corresponds to the global minimum. Our approach here is quite similar. The same four steps appear again below, but with minor changes. (New additions are in bold.) How To Find Global Maximum and Global Minimum of a One Two-Variable Function on an interval [a, b] a region R 1. Take the partial derivative of the function, set it equal to 0 and solve for x and y. These are critical points of the function. 2. Classify the critical points that are in [a, b] in R found in (1) as a local max/local min/saddle pt. or neither. (We ignore any other critical points.) 3. Find the y-values z-values of the critical points in (2), as well as the y-values of the endpoints extreme points (critical points and the corners, if they exist) on the boundary. 4. The largest y-value(s) z-value(s) corresponds to the global maximum and the smallest y-value(s) z-value(s) corresponds to the global minimum. 2There are different ways to find the critical points on the boundary. We shall keep our focus limited to fairly straightforward situations, though. If the region is square, i.e. R = [a, b] μ [c, d] (this says a § x § b and c § y § d), then we can find the points by substituting in the boundary into the function. This will reduce the function f(x, y) to just a function of either x or y. Now, we can follow apply the steps for a one-variable function to find critical points. (That is, follow steps (1) and (2).) Note: The same technique works if the region R is a triangle as well. At this point, we turn our attention to solving a variety of problems. Example 1: Find the global maximum and minimum values of f(x, y) = x2 + y2 on the region R defined by triangle with vertices at (0, 0), (2, 0) and (0, 4). Solution: We begin by finding any critical points of the function inside of the specified region R. Notice that fx(x, y) = 2x and fy(x, y) = 2y. Setting these equations equal to 0, we have that x = 0 and y = 0. Thus, we have the critical point (0, 0). Now, we could go on to use the second-derivative test to classify the critical point, but we can save some time by noticing that f(x, y) is an upwards pointing parabolloid whose smallest value will be at the origin. Thus, we have that (0, 0) corresponds to a local minimum whose z-value is f(0, 0) = 0. Next, we check the boundary of R. To do this, it helps to draw a picture of the region, since we will be breaking the boundary up into different pieces. xy24L1L2L3 Figure 1: The region R 3Now, we shall look for critical points on each of the four line segments that define the region R. We begin with L1. First off, consider the line L1. It passes through the points (2, 0) and (0, 4). Thus, the equation of the line is given by y = –2x + 4, for 0 § x § 2. Our function f(x, y) reduces to f(x, –2x + 4) = x2 + (–2x + 4)2 = 5x2 – 16x + 16, for 0 § x § 2. We look for critical points just like we did back in one-variable calculus. Taking the derivative of this function, we have 10x – 16. Setting it equal to 0, we see that x = 8/5. The corresponding y-value is –2(8/5) + 4 = 4/5. So, we end up with another critical point, namely (8/5, 4/5). Its corresponding z-value is