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Unformatted text preview: SOLUTIONS 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 ) = ∞ . 1 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 . 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 we find the critical point y = 0 . The corre sponding 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. Computing the derivative and setting it to we find 2 x 2 = 0 = ⇒ x = 1 . This gives the corner (1 , 0) as before. • the line y = 2 : f ( x, 2) = x 2 + 2 x + 4 with critical point x = 1 which is again a corner. Finally, we check the four corners ( 1 , 0) , (1 , 0) , ( 1 , 2) , (1 , 2) . The values of the function f are f ( 1 , 0) = 3 , f (1 , 0) = 1 , f ( 1 , 2) = 3 , f (1 , 2) = 7 . From the boxed values we select the lowest and the highest to find the global min and global max....
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This note was uploaded on 06/06/2011 for the course MATH 10C taught by Professor Hohnhold during the Fall '07 term at UCSD.
 Fall '07
 Hohnhold
 Critical Point, Derivative

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