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

# Calculus: Multivariable

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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 > 0 and f xx = 2 > 0 , 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 > 0 and f xx < 0 . 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 < 0 , so f has a saddle point at (0 , - 5) . When x = 2 , we have D = 12 > 0 and f xx = 6 > 0 , 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 < 0 , so f has a saddle point at (1 , - 1) and ( - 1 , 1) . At (1 , 1) , we have D = 36 > 0 and f xx = 6 > 0 , so f has a local minimum at (1 , 1) . At ( - 1 , - 1) , we have D = 36 > 0 and f xx = - 6 < 0 , so f has a local maximum at ( - 1 , - 1) .

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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 > 0 and f xx < 0 , so f has a local maximum. At the point (0 , 1) , we have D < 0 , so f has a saddle point. At the point (2 , - 1) , we have D < 0 , so f has a saddle point. At the point (2 , 1) , we have D > 0 and f xx > 0 , so f has a local minimum. 7. 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 - 12 y = 0 shows that x = - 1 or x = 1 and y = 0 or y = 4 . There are four critical points: ( - 1 , 0) , (1 , 0) , ( - 1 , 4) , and (1 , 4) . We have D = ( f xx )( f yy ) - ( f xy ) 2 = (6 x )(6 y - 12) - (0) 2 = (6 x )(6 y - 12) . At critical point ( - 1 , 0) , we have D > 0 and f xx < 0 , so f has a local maximum at ( - 1 , 0) . At critical point (1 , 0) , we have D < 0 , so f has a saddle point at (1 , 0) . At critical point ( - 1 , 4) , we have D < 0 , so f has a saddle point at ( - 1 , 4) . At critical point (1 , 4) , we have D > 0 and f xx > 0 , so f has a local minimum at (1 , 4) . 8. Find the critical point(s) by setting f x = ( xy + 1) + ( x + y ) · y = y 2 + 2 xy + 1 = 0 , f y = ( xy + 1) + ( x + y ) · x = x 2 + 2 xy + 1 = 0 , then we get x 2 = y 2 , and so x = y or x = - y .
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