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a5_soln

# a5_soln - Math 237 Assignment 5 Solutions 1 Find and...

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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 ) = x 2 y- 2 xy 2 + 3 xy + 4. Solution: We have 0 = f x = 2 xy- 2 y 2 + 3 y = y (2 x- 2 y + 3) (1) 0 = f y = x 2- 4 xy + 3 x = x ( x- 4 y + 3) (2) From (2) we get x = 0 or x = 4 y- 3. If x = 0, then (1) gives 0 = y (- 2 y + 3) so y = 0 or y = 3 2 . Hence, we get critical points (0 , 0) and (0 , 3 2 ). If x = 4 y- 3 then (1) gives 0 = y (8 y- 6- 2 y + 3) = y (6 y- 3) so y = 0 or y = 1 2 . This gives critical points (- 3 , 0) and (- 1 , 1 2 ). To classify the critical points we find that Hf ( x, y ) = 2 y 2 x- 4 y + 3 2 x- 4 y + 3- 4 x . Hf (0 , 0) = 0 3 3 0 . Then det Hf (0 , 0) =- 9 so Hf (0 , 0) is indefinite and hence (0 , 0) is a saddle point and the level curves are hyperbola. Hf (0 , 3 / 2) = 3- 3- 3 . Then det Hf (0 , 3 / 2) =- 9 so Hf (0 , 3 / 2) is indefinite and hence (0 , 3 / 2) is a saddle point and the level curves are hyperbola. Hf (- 3 , 0) =- 3- 3 12 . Then det Hf (- 3 , 0) =- 9 so Hf (- 3 , 0) is indefinite and hence (- 3 , 0) is a saddle point and the level curves are hyperbola. Hf (- 1 , 1 / 2) = 1- 1- 1 4 . Then det Hf (- 1 , 1 / 2) = 3 and f xx &amp;gt; 0 so Hf (- 1 , 1 / 2) is positive definite and hence (- 1 , 1 / 2) is a local minimum and the level curves are ellipses. b) g ( x, y ) = xye 2 x +3 y . Solution: We have 0 = g x = ye 2 x +3 y + 2 xye 2 x +3 y = ye 2 x +3 y (1 + 2 x ) (3) 0 = g y = xe 2 x +3 y + 3 xye 2 x +3 y = xe 2 x +3 y (1 + 3 y ) (4) From (3) we get y = 0 or x =- 1 2 and from (4) we get x = 0 or y =- 1 / 3. Hence, we get the critical points (0 , 0), (- 1 2 ,- 1 3 ). 2 We have Hg ( x, y ) = 4 ye 2 x +3 y + 4 xye 2 x +3 y e 2 x +3 y + 3 ye 2 x +3 y + 2 xe 2 x +3 y + 6 xye 2 x +3 y e 2 x +3 y + 3 ye 2 x +3 y + 2 xe 2 x +3 y + 6 xye 2 x +3 y 6 xe 2 x +3 y + 9 xye 2 x +3 y Hence, Hg (0 , 0) = 0 1 1 0 so det Hg (0 , 0) =- 1 and so (0 , 0) is a sadlle point and the level curves of g near (0 , 0) are hyperbola. Hg (- 1 2 ,- 1 3 ) =- 2 3 e- 2- 3 2 e- 2 , so det Hg (- 1 2 ,- 1 3 ) &amp;gt; 0 and g xx &amp;lt; 0 so (- 1 2 ,- 1 3 ) is a local maximum and the level curves of g near (- 1 2 ,- 1 3 ) are ellipses....
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a5_soln - Math 237 Assignment 5 Solutions 1 Find and...

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