Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch15 solutions

Calculus: Multivariable

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Unformatted text preview: 15.1 SOLUTIONS 1039 CHAPTER FIFTEEN Solutions for Section 15.1 Exercises 1. The point A is not a critical point and the contour lines look like parallel lines. The point B is a critical point and is a local maximum; the point C is a saddle point. 2. To find the critical points, we solve f x = 0 and f y = 0 for x and y . Solving f x = 2 x- 2 y = 0 , f y =- 2 x + 6 y- 8 = 0 . We see from the first equation that x = y . Substituting this into the second equation shows that y = 2 . The only critical point is (2 , 2) . We have D = ( f xx )( f yy )- ( f xy ) 2 = (2)(6)- (- 2) 2 = 8 . Since D > and f xx = 2 > , the function f has a local minimum at the point (2 , 2) . 3. The partial derivatives are f x =- 6 x- 4 + 2 y and f y = 2 x- 10 y + 48 . Set f x = 0 and f y = 0 to find the critical point, thus 2 y- 6 x = 4 and 10 y- 2 x = 48 . Now, solve these equations simultaneously to obtain x = 1 and y = 5 . Since f xx =- 6 , f yy =- 10 and f xy = 2 for all ( x, y ) , at (1 , 5) the discriminant D = (- 6)(- 10)- (2) 2 = 56 > and f xx < . Thus f ( x, y ) has a local maximum value at (1 , 5) . 4. To find the critical points, we solve f x = 0 and f y = 0 for x and y . Solving f x = 3 x 2- 6 x = 0 f y = 2 y + 10 = 0 shows that x = 0 or x = 2 and y =- 5 . There are two critical points: (0 ,- 5) and (2 ,- 5) . We have D = ( f xx )( f yy )- ( f xy ) 2 = (6 x- 6)(2)- (0) 2 = 12 x- 12 . When x = 0 , we have D =- 12 < , so f has a saddle point at (0 ,- 5) . When x = 2 , we have D = 12 > and f xx = 6 > , so f has a local minimum at (2 ,- 5) . 5. To find the critical points, we solve f x = 0 and f y = 0 for x and y . Solving f x = 3 x 2- 3 = 0 f y = 3 y 2- 3 = 0 shows that x = 1 and y = 1 . There are four critical points: (1 , 1) , (- 1 , 1) , (1 ,- 1) and (- 1 ,- 1) . We have D = ( f xx )( f yy )- ( f xy ) 2 = (6 x )(6 y )- (0) 2 = 36 xy. At the points (1 ,- 1) and (- 1 , 1) , we have D =- 36 < , so f has a saddle point at (1 ,- 1) and (- 1 , 1) . At (1 , 1) , we have D = 36 > and f xx = 6 > , so f has a local minimum at (1 , 1) . At (- 1 ,- 1) , we have D = 36 > and f xx =- 6 < , so f has a local maximum at (- 1 ,- 1) . 1040 Chapter Fifteen /SOLUTIONS 6. To find the critical points, we solve f x = 0 and f y = 0 for x and y . Solving f x = 3 x 2- 6 x = 0 f y = 3 y 2- 3 = 0 shows that x = 0 or x = 2 and y =- 1 or y = 1 . There are four critical points: (0 ,- 1) , (0 , 1) , (2 ,- 1) , and (2 , 1) . We have D = ( f xx )( f yy )- ( f xy ) 2 = (6 x- 6)(6 y )- (0) 2 = (6 x- 6)(6 y ) . At the point (0 ,- 1) , we have D > and f xx < , so f has a local maximum. At the point (0 , 1) , we have D < , so f has a saddle point. At the point (2 ,- 1) , we have D < , so f has a saddle point....
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Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch15 solutions

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