11.7 Maximum and Minimum Values

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1 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, i.e., local and global maxima and minima. We say that f(x, y) has a global maximum at a point (a, b) of its domain D f if f(x, y) f(a, b) for all points (x, y) in D f. That is, f(a, b) is the largest value of f in D f. We say that f(x, y) has a global minimum at a point (a, b) of its domain D f if f(a, b) f(x, y) for all points (x, y) in D f. That is, f(a, b) is the smallest value of f in D f. We say that f(x, y) has a local maximum at a point (a, b) of its domain D f if there is an R > 0 such that f(x, y) f(a, b) for all points (x, y) in D f satisfying (x a) 2 + (y b) 2 < R 2. We say that f(x, y) has a local minimum at a point (a, b) of its domain D f if there is an R > 0 such that f(a, b) f(x, y) for all points (x, y) in D f satisfying (x a) 2 + (y b) 2 < R 2. Collectively, local maxima and local minima are called local extrema. Similar definition for global extrema. Figure provides an example of a function with local and global extrema. Figure Recall that for single-variable functions y = f(x), if x = c is a local maximum or a minimum point, then either f (c) = 0 or f (c) does not exist. A point (c, f(c)) such f (c) = 0 or f (c) does not exist is called a critical point. Thus, the recipe for finding a maximum or a minimum point is to locate critical points. Something similar happens for functions of two variables. Points where the gradient is either zero or undefined are called critical points of a function f(x, y). We next show that local extrema are critical points. 1

Theorem 11.7.1 If f has a local maximum or a local minimum at a non-boundary point (a, b) in its domain then f(a, b) = 0. That is, (a, b) is a critical point. Proof.

2 Theorem If f has a local maximum or a local minimum at a non-boundary point (a, b) in its domain then f(a, b) = 0. That is, (a, b) is a critical point. Proof. Suppose f has a local extremum at a point (a, b). Define g(x) = f(x, b). Then g(x) has a local extremum at x = a so that g (a) = 0 = f x (a, b). Likewise, the function G(y) = f(a, y) has a local extremum at y = b so that G (b) = 0 = f y (a, b). Hence, f(a, b) = 0 Recall that the tangent plane to the surface z = f(x, y) is z = f(a, b) + f x (a, b)(x a) + f y (a, b)(y b) so from the above theorem this equation becomes z = f(a, b). That is, the tangent plane at a local extremum is horizontal. Just as the vanishing of the first derivative of a function in one variable does not guarantee a maximum or a minimum, the vanishing of the gradient does not guarantee a local extremum either. That is, the converse to Theorem is not true in general. Example Locate and classify the critical points of f(x, y) = x 2 y 2. The gradient of f is given by f(x, y) = 2x i 2y j. We see that f x (x, y) = 2x and f y (x, y) = 2y are simultaneously zero at (0, 0). Therefore, (0, 0) is a critical point and a possible extremum. The graph of f(x, y) shown in Figure indicates that (0, 0) is neither a local maximum nor a local minimum. Such a point will be called a saddle point Figure We need to be able to determine whether or not a function has an extreme value at a critical point. The following test is analogous to the Second Derivative Test 2

3 for functions of one variable. The Second Derivative Test Let (a, b) be a point in the domain of f such that f x (a, b) = f y (a, b) = 0. Furthermore, let D(a, b) = f xx f xy = f xx(a, b)f yy (a, b) [f xy (a, b)] 2. f yx f yy 1. If D > 0 and f xx (a, b) > 0, then f(x, y) has a relative minimum at (a, b). 2. If D > 0 and f xx (a, b) < 0, then f(x, y) has a relative maximum at (a, b). 3. If D < 0 then f(x, y)) has a saddle point at (a, b). Example Find the local extrema and saddle points of the function f(x, y) = 1 3 x3 3x 2 + y2 4 + xy + 13x y + 2. we first find the critical points for this function. This gives us: f x (x, y) =x 2 6x + y + 13 = 0 f y (x, y) = y 2 + x 1 = 0. From the second equation we find y = 2 2x. Substituting this into the first equation we find x 2 8x + 15 = (x 3)(x 5) = 0. Thus, x = 3 and x = 5 so that the critical points are (3, 4) and (5, 8). On the other hand, we have f xx (x, y) = 2x 6, f yy (x, y) = 1 2, and f xy(x, y) = 1. Hence, D(3, 4) = 1 < 0 so (3, 4) is a saddle point. Similarly, D(5, 8) = 2 1 = 1 > 0 and f xx (5, 8) = 4 > 0 so that (5, 8) is a local minimum Example Find the local extrema and saddle points of the function The partial derivatives give f(x, y) = x 3 + y 5 3x 10y + 4. f x (x, y) =3x 2 3 = 0 f y (x, y) =5y 4 10 = 0. Solving each equation we find x = ±1 and y = ± 4 2. Thus, the critical points are (1, 4 2), (1, 4 2), ( 1, 4 2), ( 1, 4 2).The discriminant is D(x, y) = f xx (x, y)f yy (x, y) [f xy (x, y)] 2 = 120xy 3. 3

4 Since D(1, 4 2) = and f xx (1, 4 2) = 6 > 0, (1, 4 2) is a local minimum. Since D(1, 4 2) = < 0, (1, 4 2) is a saddle point. Since D( 1, 4 2) = < 0, ( 1, 4 2) is a saddle point. Finally, since D( 1, 4 2) = > 0 and f xx ( 1, 4 2) = 6 < 0, ( 1, 4 2) is a local maximum The second derivative test discussed above, did not cover the case D = 0. As illustrated in the example below, the second derivative test is inconclusive in this case. That is one cannot classify the critical point. It can be either a local maximum, a local minimum or a saddle point. Example Let f(x, y) = x 4 + y 4, g(x, y) = x 4 y 4, and h(x, y) = x 4 y 4. Show that D(0, 0) = 0 for each function. Classify the critical point (0, 0) for each function. Note that f x (0, 0) = f y (0, 0) = 0 so that f(x, y) has a critical point at (0, 0). Since f xx (x, y) = 12x 2, f yy (x, y) = 12y 2 and f xy (x, y) = 0, we have D(0, 0) = f xx (0, 0)f yy (0, 0) [f xy (0, 0)] 2 = 0. But the smallest value of f(x, y) occurs at (0, 0) so that f(x, y) has a local and global minimum at (0, 0) with D(0, 0) = 0. Similarly, g x (0, 0) = g y (0, 0) = 0 so that (0, 0) is a critical point of g. Moreover, g xx (x, y) = 12x 2, g yy (x, y) = 12y 2 and g xy (x, y) = 0, we have D(0, 0) = g xx (0, 0)g yy (0, 0) [g xy (0, 0)] 2 = 0. Since g(x, y) 0, the largest value occurs at (0, 0). That is, g has a local and global maximum at (0, 0) with D(0, 0) = 0. Finally, we have h x (0, 0) = h y (0, 0) = 0 so that (0, 0) is a critical point of h. Since h xx (x, y) = 12x 2, h yy (x, y) = 12y 2 and h xy (x, y) = 0, we have D(0, 0) = h xx (0, 0)h yy (0, 0) [h xy (0, 0)] 0 = 0. However, h(0, 0) = 0, z = h(x, 0) = x 4 > 0 and z = h(0, y) = y 4 < 0. Hence, (0, 0) is a saddle point with D(0, 0) = 0 Example Find the shortest distance from the point (1, 0, 2) to the plane x + 2y + z = 4. Let d be the distance from (1, 0, 2) to any point (x, y, z) on the plane x + 2y + z = 4. By the distance formula, We have d = (x 1) 2 + y 2 + (z + 2) 2 = (x 1) 2 + y 2 + (6 x 2y) 2. 2x + 2y 7 f x (x, y) = d 2x + 5y 12 f y (x, y) =. d Solving 2x + 2y 7 = 0 and x + 5y 12 = 0 simultaneously gives x = 11 6 and y = so that ( 6, 5 3 ) is the only critical point of f. An absolute minimum exists (since there is a minimum distance from the point to the plane) and it must occur at a critical point so the shortest distance occurs when x = 11 6 and y = 5 3, 4

5 for which d = 5 6 Absolute Extrema In real life, one is most likely interested in finding the places at which the largest and smallest values of a function f occur in its domain. We recall the reader that a point (a, b) in the domain of f(x, y) is called an absolute or global maximum if f(x, y) f(a, b) for all points in the domain of f. If f(a, b) f(x, y) for all points in the domain of f then f(x, y) has an absolute or global minimun at (a, b). Optimization typically refers to finding the global maximum or minimum of a function. If the domain of f is the entire xy plane then we have an unconstrained optimization; if the domain of f is not the entire xy plane then we have a constrained optimization. Example (Unconstrained Optimization) Consider the function f(x, y) = x 2 (y + 1) 3 + y 2. Find the global extrema of f, if they exist. The first partials give f x (x, y) =2x(y + 1) 3 = 0 f y (x, y) =3x 2 (y + 1) 2 + 2y = 0. This implies that the only critical point is (0, 0). Finding second partials we have f xx (x, y) =2(y + 1) 3 f xx (0, 0) =2 f yy (x, y) =6x 2 (y + 1) + 2 f yy (0, 0) =2 f xy (x, y) =6x(y + 1) 2 f xy (0, 0) =0. Since D = f xx (0, 0)f yy (0, 0) f xy (0, 0) 2 = 4 > 0 and f xx (0, 0) = 2 > 0, the point (0, 0) is a local minimum. Since f( 3, 2) = 5 < f(0, 0) = 0 the point (0, 0)is not a global minimum. Thus, f has no global extrema Like functions in one variable, a function f(x, y) can have both a global maximum and a global minimum; a global maximum but no global minima; a global minimum but no global maxima; or none. So are there conditions that guarantee that a function has a global maximum and global minimum? In single variable calculus we saw that a function f(x) continuous on a closed (i.e., including the endpoints) and bounded (i.e. of finite length) interval has both a global maximum and a global minimum. A similar result is true for functions of two variables. However, we need to define what we mean by bounded and closed in 2D case. 5

6 A closed set in IR 2 is one which contains its boundary and with no holes in its interior. For example, the disk x 2 + y 2 1 is a closed set whereas x 2 + y 2 < 1 is not since the boundary, which is the circumference of the circle x 2 + y 2 = 1, is not included. Similarly, 0 < x 2 + y 2 1 is not closed since it has a hole at the origin. A bounded set in IR 2 is one that can be contained in a disk (x a) 2 + (y b) 2 < R. Using these definitions, we have the following theorem for multivariable functions: Theorem ( Extreme Value Theorem for Multivariable Functions) If f is a continuous function on a closed and bounded set D in IR 2 then f has a global maximum and a global minimum in D. We note that if f is not continuous or the set D is not closed or bounded, then there is no guarantee that f will have a global maximum or minimum. For example, the plane f(x, y) = x + y 1 is continuous in the entire plane but does not have global extrema since the set is not bounded. Just as in the case of single variable functions, one can find the global extrema by doing the following: Step 1. Find the values of f at the critical points of f in D. Step 2. Find the extreme values of f on the boundary of D. Step 3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. Example Find the absolute extrema of the function f(x, y) = x + y xy in the closed triangle with vertices at (0, 0), (0, 2) and (4, 0). Since D is closed and bounded and f is continuous on D, Theorem guarantees that f has global extrema in D. Step 1. The critical points are solutions to the equations f x (x, y) = 0 and f y (x, y) = 0. That is, 1 y = 0 and 1 x = 0. The only critical point is (1, 1) and f(1, 1) = 1. Step 2. Along the line from (0, 0) to (0, 2), the function f(0, y) = y has a maximum value of 2 at (0, 2) and a minimum value of 0 at (0, 0). Along the line from (0, 0) to (4, 0) the function f(x, 0) = x has a maximum value of 4 at (4, 0) and a minimum of 0 at (0, 0). Along the line from (4, 0) to (0, 2), i.e., y = 2 x 2, we have f(x, y) = f(x, 2 x 2 ) = ( ) 1 2 x for 0 x 4, a quadratic function with minimum at x = 3 2, where f( 3 2, 5 4 ) = 7 8 and a maximum at x = 4, where f(4, 0) = 4. Step 3. The absolute maximum of f on D is f(4, 0) = 4 and the absolute minimum is f(0, 0) = 0 6

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

Maxima and Minima. Terminology note: Do not confuse the maximum f(a, b) (a number) with the point (a, b) where the maximum occurs. 10-11-2010 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.