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141Bquiz6csols

141Bquiz6csols - 1 MATH 141B QUIZ#6 Solutions Date Tues...

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1 MATH 141B QUIZ#6 Solutions Date: Tues March 4, 2008 1. f x = 2 2 x = 0 when x = 1 and f y = 4 8 y = 0 when y = 1 / 2. We need to test the point ( 1 , 1 / 2). f xx = 2 , f yy = 8 and f xy = 0 . Therefore, D = ( 2)( 8) 0 2 = 16 > 0. Finally, since f xx < 0 we conclude that at local maximum occurs at ( 1 , 1 / 2). The actual maximum value is f ( 1 , 1 / 2) = 9 2( 1) + 4(1 / 2) ( 1) 2 4(1 / 2) 2 = 11 . Therefore, D) is the answer. 2. f x = 4 x 3 4 y = 0 and f y = 4 y 3 4 x = 0. Dividing these by 4 gives x 3 = y and y 3 = x . Substituting one into the other, gives ( x 3 ) 3 = x . Therefore, x 9 x = 0 , or x ( x 8 1) . Now factoring repeatedly (several difference of squares), we get x ( x 1)( x +1)( x 2 +1)( x 4 +1) = 0 . Therefore, x = 0 , ± 1 . But y = x 3 from above, so we get three points to test: (0 , 0) , ( 1 , 1) and (1 , 1) . f xx = 12 x 2 , f yy = 12 y 2 and f xy = 4 . Therefore, At (0 , 0), D = (0)(0) ( 4) 2 = 16 < 0, which means that a saddle point occurs here. At (1 , 1), D = (12)(12) ( 4) 2 = 144 16 > 0, which means that a local minimum occurs here.

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