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anonymous
 5 years ago
2x+4=lambda(2x)
2y4=lambda(2y)
(x^2)+(y^2)=1
Solve for x and y
anonymous
 5 years ago
2x+4=lambda(2x) 2y4=lambda(2y) (x^2)+(y^2)=1 Solve for x and y

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yay 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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