11Breview2sol

# 11Breview2sol - econ 11b ucsc ams 11b Review Questions 2...

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Unformatted text preview: econ 11b ucsc ams 11b Review Questions 2 Solutions 1. f (10 , 7) = (64) 1 / 2 = 8 ; f x = 5 2 (5 x + 2 y )- 1 / 2 = ⇒ f x (10 , 7) = 5 16 ; f y = (5 x + 2 y )- 1 / 2 = ⇒ f y (10 , 7) = 1 8 ; f xx =- 25 4 (5 x + 2 y )- 3 / 2 = ⇒ f xx (10 , 7) =- 25 2048 ; f xy =- 5 2 (5 x + 2 y )- 3 / 2 = ⇒ f xy (10 , 7) =- 5 1024 ; f yy =- (5 x + 2 y )- 3 / 2 = ⇒ f yy (10 , 7) =- 1 512 ; The quadratic Taylor polynomial for f ( x,y ) = √ 5 x + 2 y centered at (10 , 7): T 2 ( x,y ) = 8+ 5 16 ( x- 10)+ 1 8 ( y- 7)- 25 4096 ( x- 10) 2- 5 1024 ( x- 10)( y- 7)- 1 1024 ( y- 7) 2 . 2. Find the critical points of the functions below, and classify their critical values (as relative minimum, relative maximum, or neither) using the second derivative test. a. f ( x,y ) = 3 x 2- 12 xy + 19 y 2- 2 x- 4 y + 5 (i) Critical points: f x = 6 x- 12 y- 2 = 0 f y =- 12 x + 38 y- 4 = 0 Now, f x = 0 = ⇒ 6 x = 12 y +2, so 12 x = 24 y +4. Plugging this into the second equation gives = ⇒ - (24 y + 4) + 38 y- 4 = 0 = ⇒ 14 y- 8 = 0 = ⇒ y = 4 7 = ⇒ x = 31 21 . So there is one critical point, ( 31 21 , 4 7 ) . (ii) Second derivative test: f xx = 6 f yy = 38 f xy =- 12 = ⇒ D = 6 · 38- 144 = 84 > . Since D > 0 and f xx > 0, it follows that f ( 31 21 , 4 7 ) = 50 21 is a relative minimum value. (In fact, since the second derivatives are all constant, this is the absolute minimum value.) b. g ( s,t ) = s 3 + 3 t 2 + 12 st + 2 (i) Critical points: g s = 3 s 2 + 12 t = 0 g t = 6 t + 12 s = 0 Now, g t = 0 = ⇒ t =- 2 s , and plugging this into the first equation gives 3 s 2- 24 s = 0 = ⇒ 3 s ( s...
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## This note was uploaded on 09/16/2011 for the course ECON 11B taught by Professor Binici during the Spring '08 term at UCSC.

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11Breview2sol - econ 11b ucsc ams 11b Review Questions 2...

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