Minimizing $f(x,y,z)=y$

Question

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!

Answer 1

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.

Answer 2

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 $rank g'(0,0,0)=1 \ne 2$.