Maxima and Minima. Terminology note: Do not confuse the maximum f(a, b) (a number) with the point (a, b) where the maximum occurs.

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1 HW: 14.7: 1,5,7,13,29,33,39,51,55 Maxima and Minima In this very important chapter, we describe how to use the tools of calculus to locate the maxima and minima of a function of two variables. Definitions Definition 1. A function f(x, y) has a local maximum at the point (a, b) if f(x, y) f(a, b) for any (x, y) near (a, b). In this case the number f(a, b) is the maximum value and we say the maximum occurs at the point (a, b). Similarly, f has a local minimum at (a, b) if f(x, y) (a, b) for any (x, y) near (a, b). Terminology note: Do not confuse the maximum f(a, b) (a number) with the point (a, b) where the maximum occurs. Definition 2. The function f(x, y) has an absolute maximum at (a, b) if f(x, y) f(a, b) for every point (x, y) in the domain. The function f(x, y) has an absolute minimum at (a, b) if f(x, y) f(a, b). Every absolute maximum (minimum) is also an absolute maximum (minimum) but of course the converse is not true. Local maxima and local minima are collectively called extreme values of the function. Recall that in calculus I we call points where f (x) = 0 or where f (x) does not exist critical points. Critical points are the places where we look for extreme values of a function since if a function has a maxima or minima at a point c then c must be a critical point. More explicitly, every maxima and minima occurs at a critical point but not all critical points give maxima or minima. If you don t remember this fact, think about the example f(x) = x 3. An analogous but slightly more complicated theory holds in the two variable case. Locating Extreme Values Definition 3. A critical point of function of the function f(x, y) is a point (a, b) where either or one or both partial derivatives do not exist. f x (a, b) = f y (a, b) = 0 Theorem 1. If f has a local maximum or minimum at the point (a, b) and both partial derivatives exist at (a, b), then f x (a, b) = f y (a, b) = 0 and (a, b) is a critical point of f. Proof. If f has a maximum at (a, b) and both partial derivatives exist, then the functions g(x) = f(x, b) and h(y) = f(a, y) have maxima at a and b respectively. Hence, g (a) = h x (a, b) = 0 and h (b) = f y (a, b) = 0. The same proof works if f has a minimum. We often write the condition f x (a, b) = f y (a, b) = 0 as f(a, b) = 0. 1

2 Example 1. Let f(x, y) = 4 + (x 1) 2 + (y 3) 2. Find and classify the extreme values of f. Solution. f is a polynomial, so it is differentiable everywhere. Compute f(x, y) = 2(x 1), 2(y 3). The critical points occur when f = 0 and this happens only at (1, 3). Is this point a local maximum or minimum (or neither)? Well, you hopefully know what the graph of this function looks like, but a good way to argue is to notice that 4 + (x 1) 2 + (y 3) 2 4 for all points (x, y). Now f(1, 3) = 4 and so we must have an absolute minimum at (1, 3). Example 2. Let f(x, y) = y 2 x 2. Find and classify the extreme values of f. Solution. Again f is a polynomial, so we need only look for points where f = 0. Compute f(x, y) = 2y, 2x and hence (0, 0) is the only critical point. If we plot this function it looks like a saddle where the rider sits at the origin. In this case we can t possibly have a maximum or minimum since if you walk towards the front of the saddle from origin you will go up but if you walk to either side you will go down. To see this more carefully, notice that away from 0 on the x-axis, f(x, y) = x 2 < 0 and away from 0 on the y axis f(x, y) = y 2 > 0. So there is neither a minimum nor a maximum. The kind of arguments we had to make in the last examples will only work for very simple functions. We need a more general tool for determining if points are maxima or minima. Recall that in calculus I, we frequently make use of the second derivative test to argue that a function has a local maximum (minimum) at a critical point when the sign of the second derivative at the critical point is negative (positive). The analog here is Theorem 2 (Second Derivatives Test.). Suppose the second partials of f(x, y) are continuous in a disk centered at (a, b) and suppose that f(a, b) = 0. Let D(a, b) = f xx (a, b)f yy (a, b) [f xy (a, b)] 2 f xx (a, b) f xy (a, b) = f yx (a, b)f yy (a, b). 1. If D(a, b) > 0 and f xx (a, b) > 0 then f has a local minimum at (a, b). 2. If D(a, b) < 0 and f xx (a, b) < 0 then f has a local maximum at (a, b). 3. If D(a, b) < 0 then f is has neither a minimum nor a maximum. Such a point (a, b) is called a saddle point of f 4. If D(a, b) = 0 then the test fails and f could have a minimum, a maximum, or a saddle point. Examples Example 3. Find and classify the critical points of the function f(x, y) = x 4 + y 4 4xy

3 Solution. f is a polynomial so all of its derivatives exist. Hence the critical points occur at points where f = 0. f(x, y) = 4x 3 4y, 4y 3 4x We thus have the equations x 3 y = 0 y 3 x = 0. Substitute x 3 = y into the second equation 0 = x 9 x = x(x 8 1) = x(x 4 1)(x 4 + 1) = x(x 2 1)(x 2 + 1)(x 4 + 1). The roots are x = 0, 1, 1 and so the critical points are (0, 0), (1, 1), ( 1, 1). Now we must find D(x, y). Compute f xx = 12x 2 f xy = 4 f yy = 12y 2 f yx = f xy = 4 Then We check 12x 2 4 D(x, y) = 4 12y 2 = 144x2 y D(0, 0) = 16 < 0 D(1, 1) = 128 > 0f xx (1, 1) = 12 > 0 D( 1, 1) = 128 > 0f xx ( 1, 1) = 12 > 0 Thus, by the second derivatives test f has a saddle point at (0, 0) and local minima at ( 1, 1) and (1, 1). The value of the minimum at ( 1, 1) is f( 1, 1) = 1 and the value of the minimum at (1, 1) is f(1, 1) = 1 Example 4. Let f(x, y) = 2x 3 24xy + 16y 3. Determine and classify the critical points of f. Solution. We have f(x, y) = 6x 2 24y, 24x + 48y 2. Since f(x, y) is a polynomial, we determine the critical points by solving f(x, y) = 0. Hence which simplify to 6(x 2 4y) = 0 24(x 2y 2 ) = 0 x 2 4y = 0 x = 2y 2. Substituting the second equation into the first, (4y 4 4y = 0 y(y 3 1) = 0. 3

4 The solutions to this equation are y = 0, 1 and so the critical points are (0, 0) and (2, 1). Now we compute f xx = 12x f yy = 96y f xy = 24 D(x, y) = (12x)(96y) 24 2 = 1152xy 576 D(0, 0) = 576 D(2, 1) = 1728 Since D(0, 0) < 0, (0, 0) is a saddle point. Since since D(2, 1) > 0 and f xx (2, 1) > 0, f has a local minimum at (2, 1). Example 5. Let f(x, y) = 4x 2 4xy + y Determine the nature of the critical points of f. Solution. We have f(x, y) = 8x 4y, 4x + 2y. and hence we find the critical points by solving 8x 4y = 0 and 4x + 2y = 0. Here however, both equations reduce to y = 2x and this means every point along this line is a critical point. Now compute D(x, y) = = 0. So the test fails for every point along the line! Of course, f(x, y) factors as f(x, y) = (2x y) and if we substitute in y = 2x f(x, 2x) = 5 so every point along the line y = 2x gives a local minimum. Applications Example 6. Find the shortest distance from the point (1, 0, 2) to the plane x + 2y + z = 4. Solution. The distance is from (1, 0, 2) to any point (x, y, z) is d = (x 1) 2 + y 2 + (z + 2) 2. Now if (x, y, z) is in the plane x + 2y + z = 4, then we can write the z coordinate as Thus, we want to minimize z = 4 x 2y. d(x, y) = (x 1) 2 + y 2 + (6 x 2y) 2 Of course, minimizing distance is the same as minimizing the distance squared so we instead find the minimum of D(x, y) = (x 1) 2 + y 2 + (6 x 2y) 2 Now minimize this function using the same techniques as above. Example 7. A rectangular box without a lid is to be made from 12 m 2 of cardboard. Find the maximum volume of this box. 4

5 Solution. Let x, y, z be the length, width, and height of the box. Then volume is V = xyz. The dimensions of the box must be chosen so that we use only 12 m 2 of cardboard. Thus 2xz + 2yz + xy = 12 Solve this equation for z: z = 12 xy 2(x + y) Compute V (x, y) = 12xy x2 y 2 2(x + y) V x = y2 (12 2xy x 2 ) V 2(x + y) 2 y = x2 (12 2xy y 2 ) 2(x + y) 2. We have a critical point when both of these expressions are zero. Hence we consider the equations 12 2xy x 2 = 0 12 xy y 2 = 0 Now these imply x 2 = y 2 so that x = y since length and width are positive. Substituting y = x into either equation we obtain 12 3x 2 = 0 so that the critical point must be (22) and we can compute that z = 1. We could use the second derivatives test to argue that this point is a maximum but it is easier to argue from the point of view of the physics of the problem. In particular, there must be a maximum volume and that maximum can only occur at a critical point. The only critical point is (2, 2) and so the dimensions (2, 2, 1) maximize the volume. The Extreme Value Theorem In calculus I the extreme value theorem said that a function continuous on a closed interval [a, b] has an absolute maximum and an absolute minimum. These will occur either at the critical points or at the end points of the interval. We can state a similar result here but we must first extend the idea of closed interval to two dimensions. Definition 4. A set D is closed if the set contains all of its boundary points. A set D is bounded if D is completely contained in some disk. Theorem 3. If f is continuous on a closed, bounded set D in the plane, then f attains an absolute maximum and an absolute minimum in D. We use this theorem in the following way: given a continuous function on a closed, bounded set D we 1. Find the values of f at the critical points of f in D. 2. Find the extreme values of f along the boundary of D. 3. Rank all of these values. The smallest value is the absolute minimum and the largest is the absolute maximum. 5

6 The only complication here is that we must check the maximum value of f along the entire boundary of D. This will be very tricky unless the domain is something simple like a solid rectangle. Example 8. Find the absolute maximum and absolute minimum of the function f(x, y) = x 2 2xy +2y on the domain D = {(x, y) 0 x 3, 0 y 2}. Solution. D is a closed and bounded set (a solid rectangle with the boundary included) and f is continuous since it is a polynomial. We find and classify the critical points: f x = 2x 2y = 0 f y = 2x + 2 The only critical point is (1, 1) and f(1, 1) = 1. Along the sides of the rectangle (draw a picture!): f(x, 0) = x 2 f(3, y) = 9 4y f(x, 2) = x 2 4x + 4 f(0, y) = 2y Since 0 x 3, the first function has a minimum at x = 0 and a maximum at x = 3. Since 0 y 2, the second function has a minimum at y = 2 and a maximum at y = 0. The third function factors as f(x, 2) = (x 2) 2 and hence has a minimum at x = 2 and a maximum at x = 0. The fourth function has a minimum at y = 0 and a maximum at y = 2. We have the following extreme values: f(0, 0) = 0 f(3, 0) = 9 f(1, 1) = 1 f(2, 2) = 0 f(0, 2) = 4. Thus the absolute minimum of f is 0 and occurs at (0, 0) and (2, 2). The absolute maximum is 9 and occurs at (3, 0). 6

11.7 Maximum and Minimum Values

Arkansas Tech University MATH 2934: Calculus III Dr. Marcel B Finan 11.7 Maximum and Minimum Values Just like functions of a single variable, functions of several variables can have local and global extrema,