Finding points that are closest and most far way from origin - using calculus

Question

As the title says i want to find $(x,y,z)\in \mathbb{R}$ that are connected with thse two "equations": $$z^2-x^2-y^2=0$$ $$x-2z-3=0$$ And are the closest or most far away from origin $(0,0,0)$ So i know i need to use Lagrange multipliers and than find maximums and minimums with a system, but i get lost at trying to get that. Normally i had a function and then some "sub"functions but now i don't have a main function or i just don't notice how to solve it using calculus. I think it's solved with Lagrange multiplier, if not i am ope for suggestions.

Any help would be appreciated.

Answer 1

A first remark, if I may : minimizing the distance is the same as minimizing the square of the distance.

So, you want to minimize $$f=x^2+y^2+z^2$$ subject to the constraints $$z^2-x^2-y^2=0\qquad , \qquad x-2z-3=0$$ So, let us define $$F=x^2+y^2+z^2+\lambda(z^2-x^2-y^2)+\mu(x-2z-3)$$ and work as usual. $$F'_x=2x -2 \lambda x+\mu$$ $$F'_y=2 y-2 \lambda y$$ $$F'_z=2z+2 \lambda z-2 \mu $$ $$F'_\lambda=z^2-x^2-y^2$$ $$F'_\mu=x-2z-3$$ At the extremum, all the derivatives must be $0$.

I am sure that you can take it from here.

Answer 2

When we face such problems, the first reaction (you properly had) is to think about Lagrange multipliers.

However, in some case such as this one, the problem can be extremely simplified.

We want to minimize $$f=x^2+y^2+z^2\tag 1$$subject to the constraints $$z^2-x^2-y^2=0\tag 2$$ $$x-2z-3=0\tag 3$$ The first constraint can rewrite $$z^2=x^2+y^2\implies f=2(x^2+y^2)\tag 4$$ and the second constraint gives $z=\frac{x-3}{2}$. So $$z^2=x^2+y^2\implies \frac{1}{4} (x-3)^2=x^2+y^2\implies y^2=-\frac{3 x^2}{4}-\frac{3 x}{2}+\frac{9}{4}\tag 5$$ and since $(x,y,z)\in \mathbb{R}$, $$y^2 \geq 0 \implies -3\leq x\leq 1\tag 6$$ So, finally, we look for the extrema of the function $$f=\frac{1}{2} (x-3)^2$$ in the above range.

Simpler, isn't it ?