prelim2sol_F06

# prelim2sol_F06 - 1 Math 192 Fall 2006 Prelim 2 Solution...

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Math 192, Fall 2006 Prelim 2 Solution 1. 5 ) 2 , 1 ( ) 1 cos( 2 ) , ( , 4 ) 2 , 1 ( ) 1 sin( ) , ( , 6 ) 2 , 1 ( 2 ¡ 0 . ¡ 0 0 y y x x f x xy y x f f x y y y x f f . The linearization is therefore 8 5 4 ) 2 ( 5 ) 1 ( 4 6 ) , ( 0 . 0 . 0 . y x y x y x L . ). 1 sin( 2 ) , ( , 2 ) , ( ), 1 cos( ) , ( 0 0 0 0 x y y x f x y x f x y y x f xy yy xx R is the rectangle 1 . 1 9 . 0 d d x and 1 . 2 9 . 1 d d y . The max of ) , ( y x f xx on R is 1 . 2 ) 1 1 cos( 1 . 2 0 0 . The max of ) , ( y x f yy on R is 2 . 2 ) 1 . 1 ( 2 . The max of ) , ( y x f xy on R is 3 . 4 ) 1 . 0 ( 2 . 4 ) 1 9 . 0 sin( ) 1 . 2 ( 2 0 0 ? 0 0 . Thus M = 4.3 and 086 . 0 ) 04 . 0 ( 15 . 2 ) 1 . 0 1 . 0 ( 15 . 2 ) 2 1 ( 3 . 4 2 1 ) , ( 2 2 . d 0 . 0 d y x y x E 2. (a) The function is defined for all ( x , y ). The critical points occur where both x f and y f are zero, or where either x f or y f do not exist. . 0 2 2 y x y x f x ¡ 0 0 2 2 0 x ky f y . Substitute 0 ) 1 ( 0 ¡ k x y x . ) , ( y x f has only one critical point for 1 z k ; the critical point is (0, 0). The discriminant of ). 1 ( 4 ) 2 ( ) 2 )( 2 ( 2 2 0 0 0 0 k k f f f f xy yy xx For k > 1, 0 ! xx f and ¡ ! 0 0 2 xy yy xx f f f (0, 0) is a local minimum for k > 1. For k < 1, 0 ! xx f and ¡ ? 0 0 2 xy yy xx f f f (0, 0) is a saddle point for k < 1.

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prelim2sol_F06 - 1 Math 192 Fall 2006 Prelim 2 Solution...

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