quiz5_sol - Solution Q x y = f(1 2 f x(1 2 x-1 f y(1 2 y-2...

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Date: Feb 16, 2011 MA3160-08 Quiz 5 No Calculators! Justify all answers! Name (print): Solutions 1. (5pts.) Let f ( x , y ) = 100 - x 3 - y 3 + 3 y 2 + 3 x . Find the critical points and classify them as local maxima, local minima, saddle points, or none of these. Solution The critical points of f(x,y) are obtained from grad f ( x , y ) = ~ 0. This leads to ( f x ( x , y ) = - 3 x 2 + 3 = 0 f y ( x , y ) = - 3 y 2 + 6 y = 0 = ( x = - 1 or x = 1 y = 0 or y = 2 . Therefore, the four critical points are { ( - 1 , 0) , ( - 1 , 2) , (1 , 0) , (1 , 2) } Second derivative test: D ( x , y ) = f xx ( x , y ) f yy ( x , y ) - ( f xy ( x , y )) 2 = ( - 6 x )(6 - 6 y ) - (0) 2 = 36 x ( y - 1) . Consequently, D ( - 1 , 0) = 36 > 0 and f xx ( - 1 , 0) = 6 > 0 implies that ( - 1 , 0) is a local mimimum. D ( - 1 , 2) = - 36 < 0 implies that ( - 1 , 2) is a saddle point. D (1 , 0) = - 36 < 0 implies that (1 , 0) is a saddle point. D (1 , 2) = 36 > 0 and f xx (1 , 2) = - 6 < 0 implies that (1 , 2) is a local maximum. 2. (5pts.) Let f ( x , y ) = 1 xy . Find the quadratic Taylor polynomial Q ( x , y ) valid near (1,2).
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Unformatted text preview: Solution Q ( x , y ) = f (1 , 2) + f x (1 , 2)( x-1) + f y (1 , 2)( y-2) + 1 2 f xx (1 , 2)( x-1) 2 + f xy (1 , 2)( x-1)( y-2) + 1 2 f yy (1 , 2)( y-2) 2 f (1 , 2) = 1 2 f x (1 , 2) =-1 x 2 y ± ± ± ± ± (1 , 2) =-1 2 f y (1 , 2) =-1 xy 2 ± ± ± ± ± (1 , 2) =-1 4 f xx (1 , 2) = 2 x 3 y ± ± ± ± ± (1 , 2) = 1 f xy (1 , 2) = 1 x 2 y 2 ± ± ± ± ± (1 , 2) = 1 4 f yy (1 , 2) = 2 xy 3 ± ± ± ± ± (1 , 2) = 1 4 Therefore, the quadratic Taylor polynomial is Q ( x , y ) = 1 2-1 2 ( x-1)-1 4 ( y-2) + 1 2 ( x-1) 2 + 1 4 ( x-1)( y-2) + 1 8 ( y-2) 2...
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