final-practice-sol1

final-practice-sol1 - Math 10C - Fall 2009 - Final Exam...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 10C - Fall 2009 - Final Exam Problem 1. [ 15 points ] At what point ( x,y,z ) on the plane x + 2 y - z = 5 does the minimum of the function f ( x,y,z ) = x 2 + 2 y 2 + ( z + 1) 2 occur? Answer: Using Lagrange multipliers, we need to solve f = λ g where g ( x,y,z ) = x + 2 y - z. Computing the gradients, we conclude (2 x, 4 y, 2( z + 1)) = λ (1 , 2 , - 1) = x = λ 2 ,y = λ 2 ,z = - λ 2 - 1 . Substituting, we have x + 2 y - z = 5 = λ 2 + 2 · λ 2 + λ 2 + 1 = 5 = λ = 2 . This gives x = 1 ,y = 1 ,z = - 2 . Problem 2. [ 20 points. ] Consider the function f ( x,y ) = 3 y 2 - 2 y 3 - 3 x 2 + 6 xy. (i) [8] Find the critical points of the function. (ii) [8] Determine the nature of the critical points (local min/local max/saddle). (iii) [4] Does the function f ( x,y ) have a global minimum or a global maximum? Answer: (i) We have f x = - 6 x + 6 y = 0 = x = y f y = 6 y - 6 y 2 + 6 x = 0 = y - y 2 + x = 0 . Substituting x = y into the second equation, we obtain 2 y - y 2 = 0 = y = 0 or y = 2 . The critical points are (0 , 0) , (2 , 2) . (ii) We compute A = f xx = - 6 , B = f xy = 6 , f yy = 6 - 12 y. When x = y = 0 = AC - B 2 = ( - 6)(6) - 6 2 < 0 = (0 , 0) saddle point , x = y = 2 = AC - B 2 = ( - 6)( - 18) - 6 2 > 0 ,A < 0 = (2 , 2) local max .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(iii) We compute lim x →∞ ,y =0 f ( x,y ) = lim x →∞ - 3 x 2 = -∞ = no global min . Similarly, lim x =0 ,y →-∞ f ( x,y ) = lim y →-∞ 3 y 2 - 2 y 3 = = no global max. Problem 3. [ 13 points ] Consider the function f ( x,y ) = ln( xy 2 ) - 2 x y (i) [8] Compute the second order Taylor polynomial of f around (1 , 1) . (ii) [5] Find the tangent plane to the graph of f at the point (1 , 1 , - 2) . Answer: (i) We compute f (1 , 1) = ln 1 - 2 = - 2 f x ( x,y ) = y 2 xy 2 - 2 y = 1 x - 2 y = f x (1 , 1) = - 1 f y ( x,y ) = 2 xy xy 2 + 2 x y 2 = 2 y + 2 x y 2 = f y (1 , 1) = 4 f xx = - 1 x 2 = f xx
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/06/2011 for the course MATH 10C taught by Professor Hohnhold during the Fall '07 term at UCSD.

Page1 / 6

final-practice-sol1 - Math 10C - Fall 2009 - Final Exam...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online