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  • 5 years ago

2x+4=lambda(2x) 2y-4=lambda(2y) (x^2)+(y^2)=1 Solve for x and y

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  1. anonymous
    • 5 years ago
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    Yay for La Grange multipliers! I just took this test this morning.... So if we isolate x and y in terms of lambda, we can solve for lambda in the third equation and then find x and y using the first two equations again... So the first equation becomes: x( 1- lambda) + 2 = 0 -> x= -2/(1- lambda) The second equation becomes: y ( 1- lambda) -2 = 0 -> y= 2/(1 - lambda) Then if we use these x and y values in the third equation we get: 4/(1- lambda)^2 + 4/(1- lambda)^2 = 1 Solve for lambda and we get : lambda= 1- sqrt(8) Since we now know lambda, we can sub that value back into the equations where we isolated x and y in terms of lambda, and we get: x= -(sqrt2)/2 y= (sqrt 2)/2

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