math20c Meesue Yoo fall 09 quiz6

math20c Meesue Yoo fall 09 quiz6 - Quiz 6 Solution for MATH...

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Quiz 6 Solution for MATH 20C, 2009 1 . Let f ( x,y ) = e x - xe y . (a) Find the critical points of the function f . Ans. f x = e x - e y = 0 e x = e y , or x = y, f y = - xe y = 0 x = 0 By solving two equations x = y and x = 0, we get x = 0 ,y = 0. So, the critical point is (0 , 0). (b) Use the Second Derivative Test to analyze the critical points. Ans. f xx = e x f xy = - e y f yy = - xe y D = - xe y · e x - ( - e y ) 2 = - xe x + y - e 2 y . At the critical point (0 , 0), D (0 , 0) = - 1 < 0 Hence, (0 , 0) is a saddle point. 2 . Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2 y + 3 z = 6. Let x , y and z be the three sides of the rectangular box. Then the volume of the box is f ( x,y,z ) = xyz. Define g ( x,y,z ) by the given constraint g ( x,y,z ) = x + 2 y + 3 z = 6 . We apply the Lagrange multiplier method and find all x , y , z and λ values satisfying f = λ g, x + 2 y + 3 z = 6 . f
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This note was uploaded on 01/26/2010 for the course MATH Math 20C taught by Professor Lunasin during the Fall '08 term at UCSD.

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math20c Meesue Yoo fall 09 quiz6 - Quiz 6 Solution for MATH...

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