Unformatted text preview: solution is shown in Â±igure 29.2. CHALLENGE 29.2. We set up the Lagrangian Function L ( x, y, Î» ) = ( x âˆ’ z 1 ) 2 + ( y âˆ’ z 2 ) 2 âˆ’ Î» Ãƒ Â³ x Î± Â´ 2 + Âµ y Î² Â¶ 2 âˆ’ 1 ! , where the scalar Î» is the Lagrange multiplier For the constraint. Setting the three partial derivatives to zero yields 2( x âˆ’ z 1 ) âˆ’ 2 Î» x Î± 2 = 0 , 2( y âˆ’ z 2 ) âˆ’ 2 Î» y Î² 2 = 0 , Â³ x Î± Â´ 2 + Âµ y Î² Â¶ 2 âˆ’ 1 = 0 . We conclude that x = Î± 2 z 1 Î± 2 âˆ’ Î» , (29.1) y = Î² 2 z 2 Î² 2 âˆ’ Î» , (29.2) 179...
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 Fall '11
 Dr.Robin
 Numerical Analysis, Approximation, Derivative, Lagrangian mechanics

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