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Unformatted text preview: Problem 1. Find the critical points of the function f ( x,y ) = 2 x 3 3 x 2 y 12 x 2 3 y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Solution: Partial derivatives f x = 6 x 2 6 xy 24 x,f y = 3 x 2 6 y. To find the critical points, we solve f x = 0 = x 2 xy 4 x = 0 = x ( x y 4) = 0 = x = 0 or x y 4 = 0 f y = 0 = x 2 + 2 y = 0 . When x = 0 we find y = 0 from the second equation. In the second case, we solve the system below by substitution x y 4 = 0 ,x 2 + 2 y = 0 = x 2 + 2 x 8 = 0 = x = 2 or x = 4 = y = 2 or y = 8 . The three critical points are (0 , 0) , (2 , 2) , ( 4 , 8) . To find the nature of the critical points, we apply the second derivative test. We have A = f xx = 12 x 6 y 24 , B = f xy = 6 x, C = f yy = 6 . At the point (0 , 0) we have f xx = 24 ,f xy = 0 ,f yy = 6 = AC B 2 = ( 24)( 6) > 0 = (0 , 0)is local max . Similarly, we find (2 , 2) is a saddle point since AC B 2 = (12)( 6) ( 12) 2 = < and ( 4 , 8) is saddle since AC B 2 = ( 24)( 6) (24) 2 < . The function has no global min since lim y ,x =0 f ( x,y ) = and similarly there is no global maximum since lim x ,y =0 f ( x,y ) = . Problem 2. Determine the global max and min of the function f ( x,y ) = x 2 2 x + 2 y 2 2 y + 2 xy over the compact region 1 x 1 , y 2 . Solution: We look for the critical points in the interior: f = (2 x 2 + 2 y, 4 y 2 + 2 x ) = (0 , 0) = 2 x 2 + 2 y = 4 y 2 + 2 x = 0 = y = 0 ,x = 1 . 1 However, the point (1 , 0) is not in the interior so we discard it for now. We check the boundary. There are four lines to be considered: the line x = 1: f ( 1 ,y ) = 3 + 2 y 2 4 y. The critical points of this function of y are found by setting the derivative to zero: y (3 + 2 y 2 4 y ) = 0 = 4 y 4 = 0 = y = 1 with f ( 1 , 1) = 1 . the line x = 1: f (1 ,y ) = 2 y 2 1 . Computing the derivative and setting it to 0 we find the critical point y = 0. The corresponding point (1 , 0) is one of the corners, and we will consider it separately below. the line y = 0: f ( x, 0) = x 2 2 x....
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This note was uploaded on 03/28/2011 for the course MATH 31B taught by Professor Dragosoprea during the Winter '11 term at UCSD.
 Winter '11
 DragosOprea
 Critical Point, Derivative

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