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Unformatted text preview: Problem 1. Consider the function f ( x,y ) = 3 y 2 2 y 3 3 x 2 + 6 xy. Find the critical points of the function and determine their nature. We calculate f x = 6 x + 6 y = 0 = x = y f y = 6 y 6 y 2 + 6 x = 0 = x + y = y 2 = 2 y = y 2 = y = 0 or y = 2 . We find the second derivatives f xx = 6 , f xy = 6 , f yy = 6 12 y. We apply the second derivative test. When x = y = 0 , the Hessian is H f = 6 6 6 6 which is indefinite since the determinant is negative. The point (0 , 0) is a saddle. When y = 2 , the Hessian equals H f = 6 6 6 18 which is negative definite since the trace is negative while the determinant is positive. We found (2 , 2) is a local maximum. Problem 2. Assume that u ( x,y ) and v ( x,y ) are harmonic conjugates i.e. they satisfy the CauchyRiemann equations u x = v y , u y = v x . Show that f ( x,y ) = u ( x 2 y 2 , 2 xy ) , g ( x,y ) = v ( x 2 y 2 , 2 xy ) are also harmonic conjugates....
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This note was uploaded on 03/28/2011 for the course MATH 31B taught by Professor Dragosoprea during the Winter '11 term at UCSD.
 Winter '11
 DragosOprea
 Critical Point

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