prelim2_spring2005solutions

# prelim2_spring2005solutions - Math 192 Prelim 2 Solutions 1...

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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 > 0 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 + y ) and x = λ (2 y + x ). From the first equation we see either λ = 1 or 2 x + y = 0. If λ = 1 the second equation gives y = 0. Plugging this back into the constraint equation g we see x = ± 1. Note f (1 , 0) = f ( - 1 , 0) = 1.

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