final sol - 32A/1 WINTER QUARTER 2009 FINAL EXAM SOLUTIONS...

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32A/1 - WINTER QUARTER 2009 FINAL EXAM SOLUTIONS PROBLEM 1 VERSION 1. We are using the equation x 3 x 2 y xy 2 + y 3 =0 to put y as a function of x. Can the implicit function theorem be used ( a )(3points) at the point ( x 0 ,y 0 )=(1 , 1)? ( b x 0 0 , 1)? ( c )(4points). Assuming the implicit function theorem can be used, write an expression for y 0 ( x )= The right side of this expression will contain x, y ( x ) . SOLUTION. f ( x, y x 3 x 2 y xy 2 + y 3 , ∂f ( x, y ) ∂x =3 x 2 2 xy y 2 , ( x, y ) ∂y = x 2 2 xy +3 y 2 . ( a ) f (1 , 1)=1 1 1+1=0 , (1 , 1) = 1 2+3=0 . NO ( b ) f (1 , 1)=1+1 1 1=0 , (1 , 1) = 1+2+3=4 6 . YES ( c ) y 0 ( x ( x, y ) . ( x, y ) = 3 x 2 +2 xy ( x )+ y ( x ) 2 x 2 2 xy ( x )+3 y ( x ) 2 . PROBLEM 1 VERSION 2. We are using the equation x 3 + x 2 y xy 2 y 3 to put y as a function of x. Can the implicit function theorem be used ( a x 0 0 , 1)? ( b x 0 0 , 1)? ( c )(4points). Assuming the implicit function theorem can be used, write an expression for y 0 ( x The right side of this expression will contain x, y ( x ) . SOLUTION. f ( x, y x 3 + x 2 y xy 2 y 3 , ( x, y ) x 2 xy y 2 , ( x, y ) = x 2 2 xy 3 y 2 . 1
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( a ) f (1 , 1)=1+1 1 1=0 , ∂f (1 , 1) ∂y =1 2 3= 4 . YES ( b ) f (1 , 1)=1 1 1+1=0 , (1 , 1) =1+2 3=0 . NO ( c ) y 0 ( x )= ( x, y ) ∂x . ( x, y ) = 3 x 2 2 xy ( x )+ y ( x ) 2 x 2 2 xy ( x ) 3 y ( x ) 2 . PROBLEM 2 VERSION 1. Consider the function f ( x, y )=3 y 3 x 2 y 3 xy 2 y 3 ( a )(5points). Find all critical points of f ( x, y ) , that is, solve the system ( x, y ) =0 , ( x, y ) . Hint: the ±rst equation gives two values of y, one a function of x ; replacing each of these values in the second equation two values of x are obtained. ( b )(5points). Using the Hessian decide whether each critical point in ( a )i sa maximum, minimum or saddle point. SOLUTION. ( a )Wehave ( x, y ) = 6 xy 3 y 2 = 3 y (2 x y )=0 = y =0or y =2 x The other equation is ( x, y ) =3 3 x 2 6 xy 3 y 2 . For y =0we get 3 3 x 2 = 3(1 x 2 x =1or x = 1 . y = 2 x we get 3 3 x 2 +12 x 2 12 x 2 3 x 3 = 3(1 x 2 x x = 1 . Accordingly, (1 , 0) , ( 1 , 0) , (1 , 2) , ( 1 , 2) CRITICAL POINTS ( b f 11 ( x, y 2 f ( x, y ) 2 = 6 2 f ( x, y ) ∂x∂y = 6( x + y ) , 2 f ( x, y ) 2 = 6( x + y ) , so the Hessian is H ( x, y ¯ ¯ ¯ ¯ 6 y 6( x + y ) 6( x + y ) 6( x + y ) ¯ ¯ ¯ ¯ =36 y ( x + y ) 36( x + y ) 2 = 36( x + y )( y ( x + y )) = 36 x ( x + y ) H (1 , 0) = 36 < 0 Saddle point H ( 1 , 0) = 36 < 0 Saddle point H (1 , 2) = 36 > 0 ,f 11 (1 , 2) > 0 Minimum H ( 1 , 2) = 36 > 0 11 (1 , 2) < 0 Maximum (all local) 2
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PROBLEM 2 VERSION 2. Consider the function f ( x, y )=3 x 2 y +3 xy 2 + y 3 3 y ( a )(5points). Find all critical points of f ( x, y ) , that is, solve the system ∂f ( x, y ) ∂x =0 , ( x, y ) ∂y .
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final sol - 32A/1 WINTER QUARTER 2009 FINAL EXAM SOLUTIONS...

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