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In the problem of minimizing $f(x,y,z)=y$ over the constraint set $z=y^3 - x^2$ and $z=x^2$, I have managed to solve the problem directly and obtain that the minimum occurs at $x=0, y=0, z=0$, yielding a value of $f(0,0,0)=0$, but when I write the Lagrangian and try to solve it, I realize that it has no solutions. I am trying to understand why the Lagrange multipliers didn't work here and any insights on this would be helpful.

Edit: $\nabla f+λ_1∇g_1+λ_2∇g_2=0$ yields: $x:−2xλ_1+2xλ_2=0$ $y:1+3y^2λ_1=0$ $z:−λ_1−λ_2=0$ Which gives: $λ_1=−λ_2$ and $λ_2x=0$ but since $λ_2$ cannot be $0$, $x=0$, which gives $y^3−z=0$ and $−z=0$, giving $y=0$ and $z=0$, but $1+3y^2λ_1=0$ is not satisfied.

Thank you for your help!

Timur Lame
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  • Can you show us the details of your calculation with Lagrange multipliers? – saulspatz Mar 20 '18 at 04:48
  • Sure, here it is. – Timur Lame Mar 20 '18 at 05:01
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    Where is here? I don't see it. – saulspatz Mar 20 '18 at 05:03
  • $\nabla f+\lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 = 0$ yields:

    $x: -2x\lambda_1 +2x\lambda_2 = 0$

    $y: 1+3y^2\lambda_1=0$

    $z: -\lambda_1 - \lambda_2 = 0$

    Which gives: $\lambda_1 = -\lambda_2$ and $\lambda_2 x = 0$ but since $\lambda_2$ cannot be $0$, $x=0$, which gives $y^3-z=0$ and $-z =0$, giving $y=0$ and $z=0$, but $1+3y^2\lambda_1 = 0$ is not satisfied.

    Sorry, I pressed enter too soon above.

     – Timur Lame Mar 20 '18 at 05:08
  • Thanks, but you should really put things like this in the body of the question. Many people will look at your question and pass it by because it doesn't show any work on your part, without looking at the comments. The better your questions are, the more help you'll get. I'm going to try to figure it out from what you've said, but you may miss out with someone better at this stuff than I. – saulspatz Mar 20 '18 at 05:16
  • Thanks for the advice. I guess I'm more looking for answers to why Lagrangian multipliers can fail in general, regardless of the problem at hand, but I'll make sure to include everything in the question body next time. – Timur Lame Mar 20 '18 at 05:18

2 Answers2

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As I recall, Lagrange multipliers only find extrema in the interior of the domain. Because of $z=x^2,$ we have $z \ge 0,$ and then $y^3 = z+x^2$ gives $y\ge 0,$ so the optimum occurs on the boundary of the domain.

I've been searching online to verify my recollection about Lagrange multipliers, but I keep finding numerical examples, which doesn't help. Here I found a statement of the theorem, and it definitely requires that all the functions be defined in an open set, and only finds extrema in that open set, so that reinforces my belief.

Take a look at the statement in your textbook, and see if it bears out what I say.

saulspatz
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Let $g(x,y,z):=(z-y^3+x^2, z-x^2)$, $h(x,y,z, \lambda, \mu):=f(x,y,z)+\lambda(z-y^3+x^2)+ \mu(z-x^2)$ and $\phi:= \nabla h$.

Lagrange says: if(!) $f$ has in $(x_0,y_0,z_0) $ a local extremum under the constraints $g(x,y,z)=(0,0)$ and if(!) $rank g'(x_0,y_0,z_0)=2$, then there are $\lambda_0, \mu_0$ such that $\phi(x_0,y_0,z_0, \lambda_0, \mu_0)=0$.

In your case we have $x_0=y_0=z_0=0$, but $rank g'(0,0,0)=1 \ne 2$.

Fred
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