-2

$f(x,y)=x+y$

s.t.

$(x^2+y^2)^2=x^2-y^2$

I tried to use Lagrangian and obtained FOCs, but I had no idea how to solve them, so I also tried to do it in polar coordinates, but it seems to me that the FOCs that I obtained are also kind of hard to solve.

Maybe there exist some different approach to this problem?

Andrevv
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  • Please give more information concerning your own solution try, especially the derivation of your FOCs. – user7427029 Sep 04 '21 at 16:30
  • @user7427029 So I've set up the Lagrangian and obtained those FOCs: 1) $1- \lambda (4x^3-2x+4xy^2)$ and 2) $1- \lambda (4y^3+2y+4x^2y)$. Then I reduced it to $x(2x^2+2y^2-1)=y(2x^2+2y^2+1)$. And there I stuck. So I tried to use polar form and after the same procedure I got $r^2=(cosu+sinu)/(cosu-sinu)=cot(2u)$. – Andrevv Sep 04 '21 at 16:37

1 Answers1

0

Note that every (!) entry of the constraint function gradient must be 0 to get a critical point. In other words, both sides of your equation must equal 0. That is: $$x( 2x^2 +2y^2 -1) = 0 \\ y(2x^2 + 2y^2 +1) = 0$$

user7427029
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