math192_hw6sp06

math192_hw6sp06 - HW6 Solutions 14.7 Extreme Values and...

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Unformatted text preview: 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 > ⇒ 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 < ⇒ k > 2 or k < − 2; f will have a local minimum at (0 , 0) if 4 − k 2 > ⇒ − 2 < k < 2; the test is inconclusive if 4 − k 2 = 0 ⇒ k = ± 2...
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This homework help was uploaded on 02/05/2008 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell.

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math192_hw6sp06 - HW6 Solutions 14.7 Extreme Values and...

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