math192_hw6sp06 - HW6 Solutions 14.7 Extreme Values and...

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HW6 Solutions 14.7 Extreme Values and Saddle Points 3. critical point ( 2 3 , 4 3 ) local maximum of f ( 2 3 , 4 3 ) = 0 14. critical point (1 , 0) local minimum of f (1 , 0) = 0. 28. f x ( x, y ) = 1 x 2 + y = 0 and f y ( x, y ) = x 1 y 2 = 0 x = 1 and y = 1 the critical point is (1,1); f xx = 2 x 3 , f yy = 2 y 3 , f xy = 1; f xx (1 , 1) = 2 , f yy (1 , 1) = 2 , f xy (1 , 1) = 1 f xx f yy f 2 xy = 3 > 0 and f xx = 2 > 0 local minimum of f (1 , 1) = 3 34. Absolute maximum is 19 at (5 , 3) and the absolute minimum is 12 at (4 , 2) 41. T x ( x, y ) = 2 x 1 = 0 and T y ( x, y ) = 4 y = 0 x = 1 2 and y = 0 with T ( 1 2 , 0) = 1 4 ; on the boundary x 2 + y 2 = 1; T ( x, y ) = x 2 x + 2 for 1 x 1 T ( x, y ) = 2 x 1 = 0 x = 1 2 and y = ± 3 2 ; T ( 1 2 , 3 2 ) = 9 4 , T ( 1 2 , 3 2 ) = 9 4 , T ( 1 , 0) = 2 , and T (1 , 0) = 0 the hottest is 2 1 4 at ( 1 2 , 3 2 ) and ( 1 2 , 3 2 ); the coldest is 1 4 at ( 1 2 , 0) 46. f will have a saddle point at (0 , 0) if 4 k 2 < 0 k > 2 or k < 2; f will have a local minimum at (0 , 0) if 4 k 2 > 0 ⇒ − 2 < k < 2; the test is inconclusive if 4 k 2 = 0 k = ± 2 47. No, for example f ( x, y ) = xy has a saddle point at ( a, b ) = (0 , 0) where f x = f y = 0 54. (a) t = π 4 for 0 t π i. On the semi-ellipse: the absolute minimum is f ( 3 , 0) = 6 when t = π and the absolute maximum is f parenleftBig 3
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