# 11-14 - CHAPTER 11 Exercise 11.2 1 The derivatives are fx =...

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CHAPTER 11 Exercise 11.2 1. The derivatives are: f x = 2 x + y , f y = x + 4 y , f xx = 2 , f yy = 4 , and f xy = 1 . The fi rst-order condition requires that 2 x + y = 0 and x + 4 y = 0 . Thus we have x = y = 0 implying z = 3 (which is a minimum) 2. The derivatives are: f x = 2 x + 6 , f y = 2 y + 2 , f xx = 2 , f yy = 2 , and f xy = 0 . The fi rst-order condition requires that 2 x = 6 and 2 y = 2 . Thus we fi nd x = 3 y = 1 so that z = 10 (which is a maximum) 3. f x = 2 ax , f y = 2 by , f xx = 2 a , f yy = 2 b , and f xy = 0 . The fi rst-order condition requires that 2 ax = 0 and 2 by = 0 . Thus x = y = 0 so that z = c The second derivatives give us f xx f yy = 4 ab , and f 2 xy = 0 . Thus: (a) z is a minimum if a, b > 0 . (b) z is a maximum if a, b < 0 . (c) z gives a saddle point if a and b have opposite signs. 4. f x = 2 ¡ e 2 x 1 ¢ , f y = 4 y , f xx = 4 e 2 x , f yy = 4 , and f xy = 0 . The fi rst-order condition requires that e 2 x = 1 and 4 y = 0 . Thus x = y = 0 so that z = 4 Since f xx f yy = 4(4) exceeds f 2 xy = 0 , z = 4 is a minimum. 5. (a) And pair ( x, y ) other than (2 , 3) yields a positive z value. (b) Yes. At x = 2 and y = 3 , we fi nd f x = 4 ( x 2) 3 and f y = 4 ( y 3) 3 = 0 (c) No. At x = 2 and y = 3 , we have f xx = f yy = f xy = f yx = 0 . (d) By (11.6), d 2 z = 0 . Thus (11.9) is satis fi ed. 63

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Exercise 11.3 1. (a) q = 4 u 2 + 4 uv + 3 v 2 (b) q = 2 u 2 + 4 uv 4 v 2 (c) q = 5 x 2 + 6 xy (d) q = f xx dx 2 + 2 f xy dx dy + f yy dy 2 2. For (b): q = 2 u 2 + 4 uv 4 v 2 . For (c): q = 5 x 2 + 6 xy . Both are the same as before. 3. (a) 4 2 2 3 : 4 > 0 , 4(3) > 2 2 — positive de fi nite (b) 2 2 2 4 : 2 < 0 , 2( 4) > 2 2 — negative de fi nite (c) 5 3 3 0 : 5 > 0 , 5(0) < 3 2 — neither 4. (a) q = [ u v ] 3 2 2 7 u v (b) q = [ u v ] 1 3 . 5 3 . 5 3 u v (c) q = [ u v ] 1 4 4 31 u v (d) q = [ x y ] 2 3 3 5 x y (e) q = [ u 1 u 2 u 3 ] 3 1 2 1 5 1 2 1 4 u 1 u 2 u 3 (f) q = [ u v w ] 1 2 3 2 4 0 3 0 7 u v w 64
5. (a) 3 > 0 , 3(7) > ( 2) 2 positive de fi nite (b) 1 > 0 , 1(3) < (3 . 5) 2 neither (c) 1 < 0 , 1( 31) > 4 2 negative de fi nite (d) 2 < 0 , 2( 5) > 3 2 negative de fi nite (e) 3 > 0 , ¯ ¯ ¯ ¯ ¯ ¯ 3 1 1 5 ¯ ¯ ¯ ¯ ¯ ¯ = 14 > 0 , ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 3 1 2 1 5 1 2 1 4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 37 > 0 positive de fi nite (f) 1 < 0 , ¯ ¯ ¯ ¯ ¯ ¯ 1 2 2 4 ¯ ¯ ¯ ¯ ¯ ¯ = 0 neither (no need to check | D 3 | ) 6. (a) The characteristic equation is ¯ ¯ ¯ ¯ ¯ ¯ 4 r 2 2 3 r ¯ ¯ ¯ ¯ ¯ ¯ = r 2 7 r + 8 = 0 Its roots are r 1 , r 2 = 1 2 ¡ 7 ± 17 ¢ . Both roots being positive, u 0 Du is positive de fi nite.

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• Spring '09
• StevenJordan
• Derivatives, Convex function, first-order condition

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