Section 12.7: Maximum and Minimum Values. Definition: A function f(x, y) has a local maximum at (a, b) if f(x, y) f(a, b) for all (x, y) in

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1 Section 12.7: Maximum and Minimum Values Definition: A function f(x, y) has a local maximum at (a, b) if f(x, y) f(a, b) for all (x, y) in some disk with center (a, b). The number f(a, b) is called a local maximum value. Similarly, f(x, y) has a local minimum at (a, b) if f(x, y) f(a, b) for all (x, y) in some disk with center (a, b). The number f(a, b) is called a local minimum value. A local maximum or minimum is called a local extremum. A function f(x, y) has an absolute maximum at (a, b) if f(x, y) f(a, b) for all (x, y) in the domain of f. The number f(a, b) is called an absolute maximum value. Similarly, f(x, y) has an absolute minimum at (a, b) if f(x, y) f(a, b) for all (x, y) in the domain of f. The number f(a, b) is called an absolute minimum value. An absolute maximum or minimum is called an absolute extremum. Theorem: (Partial Derivative Criteria) If f has a local extremum at (a, b) and the first-order partial derivatives of f exist at (a, b), then f x (a, b) = f y (a, b) = 0. Definition: A point (a, b) such that f x (a, b) = f y (a, b) = 0, or one of these partial derivatives does not exist, is called a critical point of f. Example: Find all critical points of f(x, y) = x 2 + y 2 + 4x 6y + 1. f x (x, y) = 2x + 4 and f y (x, y) = 2y 6. Setting these partial derivatives equal to zero, we have x = 2 and y = 3. Thus, ( 2, 3) is the only critical point.

2 Example: Find all critical points of f(x, y) = 2x 2 + y 2 + 2xy + 2x + 2y. f x (x, y) = 4x + 2y + 2 and f y (x, y) = 2y + 2x x + 2y + 2 = 0, 2y + 2x + 2 = 0. It follows that x = 0 and y = 1. Thus, (0, 1) is the only critical point. Example: Find all critical points of f(x, y) = 2x 3 + xy 2 + 5x 2 + y 2. f x (x, y) = 6x 2 + y x and f y (x, y) = 2xy + 2y. 6x 2 + y x = 0, 2y(x + 1) = 0. It follows from the second equation that either x = 1 or y = 0. If x = 1, then the first equation becomes y 2 4 = 0 and so y = ±2. Thus, ( 1, 2) and ( 1, 2) are critical points. If y = 0, then the first equation becomes 6x x = 2x(3x + 5) = 0 and so x = 0 or x = 5/3. Thus, (0, 0) and (0, 5/3) are critical points. Definition: A critical point (a, b) of f is called a saddle point if every disk centered at (a, b) contains points (x, y) such that f(a, b) f(x, y) and points such that f(a, b) f(x, y). Theorem: (Second Partials Test) Suppose f is a function with continuous second-order derivatives and suppose (a, b) is a critical points of f. Let D = D(a, b) = f xx (a, b)f yy (a, b) [f xy (a, b)] If D > 0 and f xx (a, b) > 0, then f(a, b) is a local minimum. 2. If D > 0 and f yy (a, b) < 0, then f(a, b) is a local maximum. 3. If D < 0, then (a, b) is a saddle point.

3 Note: The function D is called the discriminant. If D = 0, the Second Partials Test gives no information. Example: Find the local extrema and saddle points of f(x, y) = 3x 2 + 8y x 12y f x (x, y) = 6x + 12 and f y (x, y) = 24y 2 24y. 6(x + 2) = 0, 24y(y 1) = 0. Thus, the critical points are ( 2, 0) and ( 2, 1). The second-order partial derivatives are Thus, the discriminant is f xx (x, y) = 6 f xy = 0 f yy = 48y 24. D(x, y) = (6)(48y 24) 0 2 = 288y 144. Since D( 2, 0) = 144 < 0, ( 2, 0) is a saddle point. f xx ( 2, 1) = 6 > 0, f( 2, 1) = 3 is a local minimum. Since D( 2, 1) = 144 > 0 and Example: Find the local extrema and saddle points of f(x, y) = x 3 3xy + y 3. f x (x, y) = 3x 2 3y and f y (x, y) = 3y 2 3x. It follows that y = x 2 and 3(x 2 y) = 0, 3(y 2 x) = 0. y 2 x = x 4 x = x(x 2 1) = x(x + 1)(x 1) = 0. This equation has three roots: x = 0, 1, 1. Thus, there are three critical points (0, 0), (1, 1), and ( 1, 1). The second-order partial derivatives are f xx (x, y) = 6x f xy = 3 f yy = 6y.

4 Thus, the discriminant is D(x, y) = (6x)(6y) ( 3) 2 = 36xy 9. Since D(0, 0) = 9 < 0, (0, 0) is a saddle point. Since D(1, 1) = 27 > 0 and f xx (1, 1) = 6 > 0, f(1, 1) = 1 is a local minimum. Since D( 1, 1) = 27 > 0 and f xx ( 1, 1) = 6 < 0, f( 1, 1) = 5 is a local maximum. Example: Find the points on the surface z 2 = xy + 1 that are closest to the origin. The distance from a point (x, y, z) on the surface to the origin is given by d = x 2 + y 2 + z 2 = x 2 + y 2 + xy + 1. d x = 2x + y 2 x 2 + y 2 + xy + 1 and d y = 2y + x 2 x 2 + y 2 + xy + 1. Setting these partial derivatives equal to zero, we have y = 2x and 2y + x = 2( 2x) + x = 3x = 0. It follows that x = y = 0 and so the points of interest are (0, 0, 1) and (0, 0, 1). Example: A cardboard box without a lid must have a volume of 23, 328 cm 3. dimensions that minimize the amount of cardboard used. Find the Let x, y, and z denote the dimensions of the box (in cm). z y x

5 The volume of the box is V = xyz = 23, 328 and so z = 23, 328/xy. Thus, the surface area of the box is given by S = xy + 2yz + 2xz = xy + S x = y x +. y and S x 2 y = x. y 2 Setting these partial derivatives equal to zero, we have y = /x 2 and so ) x = x (1 x4 y 2 = x x3 = 0. Clearly x 0 and so x = 3 = 36. Thus, y = 36 and z = 18. The dimensions of the box with smallest surface area are Definition: A closed set in R 2 is a set that contains all of its boundary points. A bounded set in R 2 is a set that is contained in some disk. Theorem: (Extreme Value Theorem) If f is continuous on a closed, bounded set D in R 2, then f attains an absolute maximum and minimum value on D. To find the absolute extrema of a continuous function f on a closed, bounded set D: 1. Find all extreme values in D. 2. Find all extreme values on the boundary of D. 3. The largest of these values is the absolute maximum and the smallest is the absolute minimum. Example: Find the absolute extrema of f(x, y) = 2+2x+2y x 2 y 2 on the closed triangular region T with vertices (0, 0), (0, 4), and (4, 0). Since f is a polynomial, it is continuous on the closed, bounded set T. The first-order partial derivatives are f x (x, y) = 2 2x and f y (x, y) = 2 2y. Thus, the only critical point in the interior of T is (1, 1) and f(1, 1) = 4. The boundary of T consists of three line segments L 1, L 2, L 3.

6 0,4 L 3 y = 4 x L 2 0,0 L 1 4,0 On L 1, we have y = 0 and f(x, 0) = 2 + 2x x 2, 0 x 2. Therefore f (x) = 2 2x = 0 and so x = 1. Thus, (1, 0) is a critical point and f(0, 0) = 2, f(1, 0) = 3, f(4, 0) = 6. On L 2, we have x = 0 and f(0, y) = 2 + 2y y 2, 0 y 2. Therefore f (y) = 2 2y = 0 and so y = 1. Thus, (0, 1) is a critical point and f(0, 0) = 2, f(0, 1) = 3, f(0, 4) = 6. On L 3, we have y = 4 x and f(x, 4 x) = 2 + 2x + 2(4 x) x 2 (4 x) 2 = 8x 2x 2 6. Therefore, f (x) = 8 4x = 0 and so x = 2. Thus, (2, 2) is a critical point and f(2, 2) = 2. The absolute maximum value is f(1, 1) = 4 and the absolute minimum value is f(4, 0) = f(0, 4) = 6.