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Using lagrange i got something like $3x = 4z = 6y$

And the constraint is $z^2 = x^2 + y^2$

Where do you get from here?

I usually get $x=y=z$, but here i got $3$ variables with different values.

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    Your next-to-last comment should be part of the question-it is what you (seem to) really want. But neither one is the function you want to maximize, they are both constraints (see the equal sign).2012-11-16

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The only solution to that system is $x=y=z=0$. Use $z=\frac{3}{4}x,y=\frac{1}{2}x$ and substitute into the constraint. The only solution is $x=0$, and the other variables follow.

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    Yes, I believe I wrote that the only solution is $x= y=z=0$?2012-11-15