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Unformatted text preview: solution is shown in igure 29.2. CHALLENGE 29.2. We set up the Lagrangian Function L ( x, y, ) = ( x z 1 ) 2 + ( y z 2 ) 2 x 2 + y 2 1 ! , where the scalar is the Lagrange multiplier For the constraint. Setting the three partial derivatives to zero yields 2( x z 1 ) 2 x 2 = 0 , 2( y z 2 ) 2 y 2 = 0 , x 2 + y 2 1 = 0 . We conclude that x = 2 z 1 2 , (29.1) y = 2 z 2 2 , (29.2) 179...
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

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