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 , f yy =4 ,and f xy =1 .T h e f rst-order condition requires that 2 x + y =0 and x y h u sw eh a v e x = y 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 f xy h e f rst-order condition requires that 2 x = 6 and 2 y = 2 h u e f nd x y so that z =10 (which is a maximum) 3. f x ax , f y by , f xx a , f yy b f xy h e f rst-order condition requires that 2 ax and 2 by h u s x = y so that z = c The second derivatives give us f xx f yy ab f 2 xy h u s : (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 ¡ e 2 x 1 ¢ , f y y , f xx e 2 x , f f xy h e f rst-order condition requires that e 2 x and 4 y h u s x = y so that z Since f xx f =4(4) exceeds f 2 xy , z is a minimum. 5. (a) And pair ( x,y ) other than (2 , 3) yields a positive z value. (b) Yes. At x and y ,we f nd f x =4( x 2) 3 and f y y 3) 3 (c) No. At x and y ,wehave f xx = f = f xy = f yx . (d) By (11.6), d 2 z . Thus (11.9) is satis f 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 uv 4 v 2 (c) q =5 x 2 +6 xy (d) q = f xx dx 2 +2 f xy dxdy + f yy dy 2 2. For (b): q = 2 u 2 uv 4 v 2 .F o r ( c ) : q x 2 xy . Both are the same as before. 3. (a) 42 23 : 4 > 0 , 4(3) > 2 2 — positive de f nite (b) 22 2 4 : 2 < 0 , 2( 4) > 2 2 — negative de f nite (c) 53 30 : 5 > 0 , 5(0) < 3 2 —ne ither 4. (a) q =[ uv ] 3 2 27 u v (b) q ] 13 . 5 3 . u v (c) q ] 14 4 31 u v (d) q xy ] 3 5 x y (e) q u 1 u 2 u 3 ] 3 12 15 1 2 u 1 u 2 u 3 (f) q uvw ] 3 2 40 7 u v w 64
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5. (a) 3 > 0 , 3(7) > ( 2) 2 — positive de f nite (b) 1 > 0 , 1(3) < (3 . 5) 2 —n e i t h e r (c) 1 < 0 , 1( 31) > 4 2 — negative de f nite (d) 2 < 0 , 2( 5) > 3 2 — negative de f nite (e) 3 > 0 , ¯ ¯ ¯ ¯ ¯ ¯ 3 1 15 ¯ ¯ ¯ ¯ ¯ ¯ =14 > 0 , ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 3 12 1 2 14 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ =37 > 0 — positive de f nite (f) 1 < 0 , ¯ ¯ ¯ ¯ ¯ ¯ 2 4 ¯ ¯ ¯ ¯ ¯ ¯ =0 — neither (no need to check | D 3 | ) 6. (a) The characteristic equation is ¯ ¯ ¯ ¯ ¯ ¯ 4 r 2 23 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 f nite. (b) The characteristic equation is r 2 +6 r +4=0 ,w ithroots r 1 , r 2 = 3 ± 5 .B o t h r o o t s being negative, u 0 Eu is negative de f nite.
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11-14 - CHAPTER 11 Exercise 11.2 1. The derivatives are: fx...

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