20Chw5 - MATH 20C graded homework 5(20 points 14.7 8(10...

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14.7 # 8 (10 points): Find the critical points of f ( x,y ) = x 3 + 2 xy - 2 y 2 - 10 x . Use the Second Derivative Test to determine whether they are local minima or maxima. Solution: The critical points of f are the points where f is not defined, or is zero. Since f is a polynomial, then f x and f y exists for all ( x,y ) in the domain of f . So we need to solve f = h 3 x 2 + 2 y - 10 , 2 x - 4 y i = h 0 , 0 i . Equating the second coordinate gives y = 1 2 x . Substituting into the equation from the first coordinate gives 3 x 2 + 2( 1 2 x ) - 10 = 0 ( x + 2)(3 x - 5) = 0 , so x = - 2 or x = 5 3 . Since y = 1 2 x , then the critical points of f are ( - 2 , - 1) and ( 5 3 , 5 6 ) . Next, calculate the discriminant. D ( x,y ) = det ± 6 x 2 2 - 4 ² = - 24 x - 4 . D ( - 2 , - 1) = 44 > 0, and f xx ( - 2 , - 1) = - 12 < 0, so f has a local maximum at ( - 2 , - 1), (and the value is f ( - 2 , - 1) = 14). D ( 5 3 , 5 6 ) = - 44 < 0, so f has a saddle point at ( 5 3 , 5 6 ). 14.8 # 16 (10 points): Find the rectangular box of maximum volume if the sum of the lengths of the edges is 300 cm. Solution : Let x,y,z denote the width, length and height of the box. We want to maximize the volume V ( x,y,z ) = xyz, subject to the constraint g ( x,y,z ) = 4( x + y + z ) = 300 . Use the Lagrange multiplier method and solve V = λ g and g = 300 simultaneously. Since
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This note was uploaded on 03/28/2011 for the course MATH 20C taught by Professor Helton during the Winter '08 term at UCSD.

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20Chw5 - MATH 20C graded homework 5(20 points 14.7 8(10...

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