F xx 00 6 k ii f y y 00 2 k 12 2 k iii f xy 00 2 k b

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f xx ( 0,0 ) =− 6 k ( 0 ) = 0 ii. f y y ( 0,0 ) = 2 k 12 ( 0 ) = 2 k iii. f xy ( 0,0 )= 2 k
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b. f xx ( 0,0 ) ∙ f y y ( 0,0 ) [ f xy ( 0,0 ) ] 2 i. 2 k ¿ ¿ 0 2 k ¿ ii. d < 0 for asaddle point : 1. When 0 is plugged in for k , the inequality does not hold true because 0 0. 2. When any nonzero value is plugged in for k, the inequality holds true because any negative square will be positive but returns to negative when multiplied by -4. Additionally, any positive squared will turn negative when multiplied by -4. 3. (− ∞, 0 ) ( 0, ) is the domain values for k. B. The function f ( x , y ) has a critical point at ( 1 3 , 1 6 ) . Use the second derivative test to demonstrate that for k = 1 2 , f has local maximum at ( 1 3 , 1 6 ) Use algebra and/or calculus techniques to justify all work. *From part A 1) First order of partial derivatives: a. f x =− 3 k x 2 + 2 ky b. f y = 2 ky 6 y 2 2 kx 2) Second order of partial derivatives: a. f xx =− 6 kx b. f y y = 2 k 12 y c. f xy = 2 k *Plug in value for k and the critical points ( 1 3 , 1 6 ) 3) Examine discriminate: a. Values for f xx ( 1 3 , 1 6 ) ,f y y ( 1 3 , 1 6 ) , k = 1 2 :
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i. f xx ( 1 3 , 1 6 ) =− 6 ( 1 2 )( 1 3 ) =− 1 ii. f y y ( 1 3 , 1 6 ) = 2 ( 1 2 ) 12 ( 1 6 ) =− 1 2 =− 3 iii. f xy ( 0,0 ) = 2 ( 1 2 ) =− 1 * f xx < 0 b. f xx ( 1 3 , 1 6 ) ∙f y y ( 1 3 , 1 6 ) [ f xy ( 1 3 , 1 6 ) ] 2 i. 1 ¿ ¿ 1 3 ¿ c. In order for a local maximum to occur at a given point (a,b), f xx < 0 AND f xx ∙f yy f xy 2 > 0 at (a,b). i. From 3a, f xx < 0 1 < 0 ii. From 3b, f xx ∙f yy f xy 2 > 0 2 > 0 Thus, there is a local maximum at the point ( 1 3 , 1 6 ) .
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