Tutorial6_sol

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,
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This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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