sprelim2 - Math 192, Fall 2006 Prelim 2 Solution 1. f (1,2)...

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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 = + = = = = 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 ) , ( + = + + = y x y x y x L . ). 1 sin( 2 ) , ( , 2 ) , ( ), 1 cos( ) , ( = = = 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 x and 1 . 2 9 . 1 y . The max of ) , ( y x f xx on R is 1 . 2 ) 1 1 cos( 1 . 2 = . 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 = < . 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 = = + + 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 2 2 = = x ky f y . Substitute 0 ) 1 ( = = k x y x . ) , ( y x f has only one critical point for 1 k ; the critical point is (0, 0). The discriminant of ). 1 ( 4 ) 2 ( ) 2 )( 2 ( 2 2 = = = k k f f f f xy yy xx For k > 1, 0 > xx f and > 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 2 xy yy xx f f f (0, 0) is a saddle point for k < 1.
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sprelim2 - Math 192, Fall 2006 Prelim 2 Solution 1. f (1,2)...

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