Lagrange Multiplier

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  • #1
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Use Lagrange Multipliers to find the maximum and minimum values of f(x,y)=x[tex]^{2}[/tex]y subject to the constraint g(x,y)=x[tex]^{2}[/tex]+y[tex]^{2}[/tex]=1.



[tex]\nabla[/tex]f=[tex]\lambda[/tex][tex]\nabla[/tex]g

[tex]\nabla[/tex]f=<2xy,x[tex]^{2}[/tex]>
[tex]\nabla[/tex]g=<2x,2y>

1: 2xy=2x[tex]\lambda[/tex] ends up being y=[tex]\lambda[/tex]
2: x[tex]^{2}[/tex]=2y[tex]\lambda[/tex] ends up being(1 into 2) x=[tex]\sqrt{2\lambda[/tex] ^{2}}[/tex]
3: x[tex]^{2}[/tex]+y[tex]^{2}[/tex]=1

1 and 2 into 3:
(2[tex]\lambda[/tex][tex]^{2}[/tex])+([tex]\lambda[/tex][tex]^{2}[/tex])=1

Do I then solve for [tex]\lambda[/tex] and x and y? Did I do the above correctly? I am going off of an example of f(x,y,z) so I'm not sure if I'm correct. Any help is greatly appreciated!
 

Answers and Replies

  • #2
68
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Be careful when dividing by a variable because x=0 could easily be a solution to this problem. Instead of trying to solve for lambda, try eliminating it. You could try multiplying the first equation by y and the second equation by x. You should get an equation for x and y, and so coupled with equation 3, you have two equations two unknowns.
 

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