hw67ans

# hw67ans - f ( x, y ) = (5 y 2 − 2) e − 2 . (And on the...

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Math 20C Prof. Fitzsimmons Answers to even-numbered exercises from sections 14.4-14.7 § 14.4 (4) 1 . 3 (20) (4 / 5 , ln(25 / 4)) ( . 8 , 1 . 83) (22) f ( 2 . 1 , 3 . 1) 4 § 14.5 (26) D a 12 / 13 , 5 / 13 A f (2 , 2) = a 2 , 2 A · a 12 / 13 , 5 / 13 A = 34 / 13 (34) Increasing, because D v f ( P ) = 12 > 0. (40) (4 / 17 , 9 / 17 , 2 / 17) and ( 4 / 17 , 9 / 17 , 2 / 17) § 14.7 (32) (a) There is one critical point at (1 , 1), and f (1 , 1) = 1. (b) Minimum: 0 at (0 , 0); maximum 4 at (0 , 2) (c) Top edge: minimum 0 at (2 , 2); maximum 4 at (0 , 2). Left edge: minimum 0 at (0 , 0); maximum 4 at (0 , 2). Right edge: minimum 0 at (2 , 2); maximum 4 at (2 , 0). (d) The maximum value of f is 4 (and it occurs at (0 , 2) and (2 , 0)); the minimum value of f is 0 (and it occurs at (0 , 0) and (2 , 2)). (38) The interior critical points are (0 , 0) , (0 , 1) , (0 , 1) , (1 , 0) , ( 1 , 0). The values of f at these points are 0, 4 e 1 , 4 e 1 , e 1 , e 1 , respectively. On the boundary circle x 2 + y 2 = 2 we have
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Unformatted text preview: f ( x, y ) = (5 y 2 − 2) e − 2 . (And on the boundary y varies from − √ 2 to √ 2.) The maximum value of f on the boundary is there-fore f (0 , ± √ 2) = 8 e − 2 ≈ 1 . 08; the minimum value of f on the boundary is f ( ± √ 2 , 0) = − 2 e − 2 ≈ − . 27. Since 4 e − 1 ≈ 1 . 47 and − e − 1 ≈ − . 37, the global maximum of f is 4 e − 1 (occurring at (0 , ± 1)) and the minimum value is − e − 1 (occurring at ( ± 1 , 0)). (42) The Lagrange multiplier equations for this problem are 2( y + z ) = λyz, 2( x + z ) = λxz, 2( x + y ) = λxy. Elimination of λ leads to 1 y + 1 z = 1 x + 1 z = 1 x + 1 y . From this it should be clear that x = y = z = V 1 / 3 ....
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## This note was uploaded on 07/04/2010 for the course PHYS PHYS 2B taught by Professor Sinha during the Spring '09 term at UCSD.

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