Minimizing $x^2+y^2+z^2$ subject to $xy -z + 1 = 0$ via Lagrange multipliers

Question

$$\begin{array}{ll} \text{minimize} & f(x,y,z) := x^2 + y^2 + z^2\\ \text{subject to} & g(x,y,z) := xy - z + 1 = 0\end{array}$$

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I tried the Lagrange multipliers method and the system resulted from has no solution. So I posted it to see if the question is wrong by itself or I'm missing something.

So I made the Lagrangian equation $L(x,y,z,λ)=x^2 + y^2 + z^2 + λ(xy -z+1)$

and then

$θL/θx = 2x + λy =0$

$θL/θy = 2y + λx =0$

$θL/θz = 2z - λ =0$

$θL/θλ = xy -z +1 =0 $

The obvious solution for that system is x=0 , y=0 , z=1 and λ=2

But solving it in an online solver for nonlinear systems of equation the answer I get is that it's unsolvable.

So my question is: What I'm doing wrong

Answer 1

Since $f$ and $g$ are polynomial, using SymPy's solve_poly_system:

>>> from sympy import *
>>> x, y, z, mu = symbols('x y z mu', real=True)
>>> L = x**2 + y**2 + z**2 + mu * (x*y - z + 1)
>>> solve_poly_system([diff(L,x), diff(L,y), diff(L,z), diff(L,mu)], x, y, z, mu)
[(0, 0, 1, 2), (-sqrt(2)*I, -sqrt(2)*I, -1, -2), (sqrt(2)*I, sqrt(2)*I, -1, -2)]

Hence, the only real solution is $(x,y,z,\mu) = (0, 0, 1, 2)$. Not very insightful, however.

Let $\mathcal L$ be the Lagrangian. Computing $\partial_x \mathcal L$, $\partial_y \mathcal L$ and $\partial_z \mathcal L$ and finding where they vanish,

$$\begin{bmatrix} 2 & \mu & 0\\ \mu & 2 & 0\\ 0 & 0 & 2\end{bmatrix} \begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \mu\end{bmatrix}$$

Note that the matrix is singular when $\mu = \pm 2$. Hence, we have three cases to consider.

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$\color{blue}{\boxed{\mu = 2}}$

The solution set is the line parameterized by

$$\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} t\\-t\\ 1\end{bmatrix}$$

and, since, $xy - z + 1 = 0$, we obtain $t = 0$ and $\color{blue}{(x,y,z) = (0,0,1)}$.

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$\color{blue}{\boxed{\mu =-2}}$

The solution set is the line parameterized by

$$\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} t\\ t\\ -1\end{bmatrix}$$

and, since, $xy - z + 1 = 0$, we obtain the equation $t^2 = -2$, which has no solution over the reals.

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$\color{blue}{\boxed{\mu \neq \pm2}}$

The solution set is the line parameterized by

$$\begin{bmatrix} x\\ y\\ z\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \frac{\mu}{2}\end{bmatrix}$$

and, since, $xy - z + 1 = 0$, we obtain $\mu = 2$, which contradicts the assumption.

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Answer 2

$$x^2+y^2+z^2=x^2+y^2+z^2+2(xy-z+1)=(x+y)^2+(z-1)^2+1\geq1.$$ The equality occurs for $x=y=0$ and $z=1$, which says that we got a minimal value.

Answer 3

This can be solved in at least two methods. First, let's solve without Lagrange, using convenient changes of variables. Let $u=x+y, v=xy$. This results in $u^2=x^2+y^2+2xy=x^2+y^2+2v$.

We now need to minimize $u^2-2v+z^2$ under the constraint $v-z+1=0$. We can rearrange this constraint to be of the form $z=1+v$ and therefore $z^2=1+2v+v^2$. Substituting this, we need to minimize $u^2+v^2$. The minimum of this is for $u=0, v=0$, which returns $x=0, y=0, z=1$.

Solving this using Lagrange:

$$L=x^2+y^2+z^2-\lambda(xy-z+1)=x^2+y^2+z^2-\lambda xy-\lambda z-\lambda$$ $$\frac{\partial L}{\partial z}=2z-\lambda\rightarrow\lambda=2z$$ $$\frac{\partial L}{\partial x}=2x-\lambda y=0\rightarrow x=\frac{\lambda y}{2}=yz$$ $$\frac{\partial L}{\partial y}=2y-\lambda x=0\rightarrow2y-2yz^2=0$$$$\rightarrow y=0, x=0\cup z=1,\lambda=2,x=y\cup z=-1,\lambda=-2,x=-y$$ We have three possible solutions to this. We will plug each into the equation for the constraint $xy-z+1=0$

If $x=0, y=0$, our constraint becomes $-z+1=0$, which has the solution $x=0, y=0, z=1$, with the value of $x^2+y^2+z^2=1$

If $z=1, x=y$, our constraint becomes $x^2-1+1=0$, which has the exact same solution

If $z=-1, x=-y$, cour constraint becomes $-y^2+1+1=0$, which has the solutions $x=\pm\sqrt{2}, y=\mp\sqrt{2}, z=-1$. The value here is $x^2+y^2+z^2=5$, which is not the minimum