Lagrange method over two constraints

Question

plane $x+y-z=-2$ intersects $z^2=x^2+y^2$

I need to use Lagrange multipliers to determine the point of intersection which is the closest to the origin.

As far as I understand, to use Lagrange I need a function $f(x,y,z)$ so that: $$f(x,y,z)=\lambda G(x,y,z)+\mu F(x,y,z)$$

But what function $f$ should I use here?

Answer 1

$f(x,y,z)=\sqrt{x^2+y^2+z^2}$ assuming that you are talking about the Euclidean minimum distance

Here you have two constraints, that means two Lagrangian multipliers namely

$\lambda_1$ and $\lambda_2$ for the constraints $x+y-z+2=0$ and $x^2+y^2-z^2=0$. One can write them all as

$$\mathcal{L}(x,y,z,\lambda_1,\lambda_2)=\sqrt{x^2+y^2+z^2}+\lambda_1(x+y-z+2)+\lambda_1(x^2+y^2-z^2)$$

Answer 2

You should solve the problem

\begin{equation} \begin{array}{c} \min \hspace{3mm} x^2 +y^2 +z^2 \\ s.t. \hspace{3mm} x+y -z +2 =0 \\ \hspace{8mm} x^2+y^2-z^2 = 0. \\ \end{array} \end{equation}

The lagrangian function is given by

$$\mathcal{L} = x^2 +y^2 +z^2 + \lambda (x+y -z +2)+\mu (x^2+y^2-z^2).$$ Hence,

$$\frac{\partial \mathcal{L}}{\partial x} = 2x +\lambda + 2x\mu= 0 $$

$$\frac{\partial \mathcal{L}}{\partial y} = 2y +\lambda + 2y\mu= 0$$

$$\frac{\partial \mathcal{L}}{\partial z} = 2z -\lambda - 2z\mu= 0 .$$

Now, I think you can solve this system.