prelim2_spring2005solutions

prelim2_spring2005solutions - Math 192, Prelim 2 April 14,...

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Unformatted text preview: Math 192, Prelim 2 April 14, 2005 Solutions 1) a) Consider f ( x,y ) =- x 2- y 2 . Then f x =- 2 x and f y =- 2 y so the only critical point is (0 , 0). Observe f xx (0 , 0) =- 2 = f yy (0 , 0) and f xy (0 , 0) = f yx (0 , 0) = 0 so D (0 , 0) = 4 > and we have a maxima. b) Consider f ( x,y ) = x 2- y 2 . Then f x = 2 x and f y =- 2 y so the only critical point is (0 , 0). Observe f xx (0 , 0) = 2, f yy (0 , 0) =- 2 and f xy (0 , 0) = f yx (0 , 0) = 0 so D (0 , 0) =- 4 < 0 and we have a saddle. 2) Since f ( x,y ) = x 2 + xy we see f x = 2 x + y and f y = x . Solving 2 x + y = 0 and x = 0 simultaneously gives that (0 , 0) is the only critical point. Note (0 , 0) is in the region consisting of points ( x,y ) satisfying x 2 + xy + y 2 ≤ 1. As f xx (0 , 0) = 2, f yy (0 , 0) = 0 and f xy (0 , 0) = f yx (0 , 0) = 1 we see D (0 , 0) =- 1 < 0 so (0 , 0) is a saddle. We now analyze the boundary given by g ( x,y ) = x 2 + xy + y 2- 1 = 0 using Lagrange multipliers. Setting ~ ∇ f = λ ~ ∇ g we see 2 x + y = λ (2 x...
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This note was uploaded on 06/01/2008 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell.

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prelim2_spring2005solutions - Math 192, Prelim 2 April 14,...

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