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Unformatted text preview: Math 237 Assignment 5 Solutions 1. Find and classify the critical points of the following functions and determine the shape of the level curves near each critical point. a) f ( x, y ) = 1 2 x 2 + 3 y 3 + 9 y 2 3 xy + 9 y 9 x . Solution: We need to solve 0 = f x = x 3 y 9 (1) 0 = f y = 9 y 2 + 18 y 3 x + 9 (2) (1) gives x = 3 y + 9. Putting this into (2) we get 0 = 9 y 2 + 18 y 3(3 y + 9) + 9 = 9 y 2 + 9 y 18 = 9( y 1)( y + 2). Therefore we get y = 1 or y = 2 which gives x = 12 and x = 3 respectively. Hence we have critical points (12 , 1) and (3 , 2). We have Hf ( x, y ) = 1 3 3 18 y + 18 . At (12 , 1) we get Hf (12 , 1) = 1 3 3 36 . Then det Hf (12 , 1) = 27 > 0 and f xx (12 , 1) > so Hf (12 , 1) is positive definite and so (12 , 1) is a local min and the level curves are ellipses. At (3 , 2) we get Hf (3 , 2) = 1 3 3 18 . Then det Hf (3 , 2) = 27 < 0 so f (3 , 2) is indefinite and so (3 , 2) is a saddle point and the level curves are hyperbola. b) g ( x, y ) = x 3 + y 3 6 y 2 3 x + 9. Solution: We need to solve 0 = g x = 3 x 2 3 = 3( x 1)( x +1), 0 = g y = 3 y 2 12 y = 3 y ( y 4). Hence, we have 4 critical points (1 , 0), ( 1 , 0), (1 , 4) and ( 1 , 4). We have Hg ( x, y ) = 6 x 6 y 12 . At (1 , 0) we have Hg (1 , 0) = 6 12 and so det g (1 , 0) = 72 < 0 and so it is indefinite and thus (1 , 0) is a saddle point and the level curves are hyperbola. At ( 1 , 0) we have Hg ( 1 , 0) = 6 12 and so det g ( 1 , 0) = 72 > 0 and g xx ( 1 , 0) < 0 so it is negative definite and thus ( 1 , 0) is a local maxium and the level curves are ellipses. At (1 , 4) we have Hg (1 , 4) = 6 0 12 and so det g (1 , 4) = 72 > 0 and g xx (1 , 4) > 0 so it is positive definite and thus (1 , 4) is a local minimum and the level curves are ellipses. At ( 1 , 4) we have Hg ( 1 , 4) = 6 12 and so det g (1 , 4) = 72 < 0 and so it is indefinite and thus (1 , 4) is a saddle point and the level curves are hyperbola. 2 c) h ( x, y ) = ( x + y )( xy + 1). Solution: We need to solve 0 = h x = xy + 1 + y ( x + y ) = 2 xy + y 2 + 1 (3) 0 = h y = xy + 1 + x ( x + y ) = 2 xy + x 2 + 1 (4) We have y 2 + 1 = 2 xy = x 2 + 1 and hence x 2 = y 2 . If x = y then (3) gives 0 = 3 y 2 + 1 which has no solutions. If x = y...
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This note was uploaded on 01/04/2012 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.
 Spring '08
 WOLCZUK
 Critical Point

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