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Tutorial6_sol - MATH1211/10-11(1/Tu6sol Department of...

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MATH1211/10-11(1)/Tu6sol/ Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010–2011, First semester) Brief Solution to Tutorial 6 1. (a) False . Counter-example: the constant function f ( x, y ) = C . At (0 , 0) (or any point in R 2 ) f has both a local (in fact global) minimum and a local (in fact global) maximum. (b) True . Suppose a is a saddle point of f . Then in any neighborhood of a there are points b and c such that f ( b ) < f ( a ) < f ( c ). Hence f could not have a local minimum/maximum at a . (c) False . Counter-example: f ( x, y ) = 2 x 2 +2 y 2 , g ( x, y ) = - x 2 - y 2 . Then f has a local minimum at (0 , 0) and g has a local maximum at (0 , 0). But (0 , 0) is not a saddle point of f + g = x 2 + y 2 . 2. To solve for critical points: ( 0 = f x = 6 x 2 - 6 y 0 = f y = - 6 x + 6 y = ( x 2 = y x = y = ( x, y ) = (0 , 0) or (1 , 1) . Also, f xx = 12 x, f xy (= f yx ) = - 6 , f yy = 6. At critical point (0 , 0), f xx = 0 and f xx f yy - f 2 xy = - 36 < 0. Hence (0 , 0) is a saddle point. At critical point (1 , 1), f xx > 0 and f xx f yy - f 2 xy = 36 > 0; hence f has a local minimum at (1 , 1). 3. Write g ( x, y, z ) = x + y - z so that the constraint becomes g ( x, y, z ) = 1. Critical points are obtained by solving f = λ g and g ( x, y, z ) = 1: 2 x = λ 2 y = λ 2 z = - λ x + y - z = 1 . The first three equations give
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