MHB Points of the surface with minimum distance to the point (3,0,0)

[mathmari]

Hey! :giggle:

We have the function $$f(x,y)=(x-3)^2+y^2+(x-y)^2$$ and I have shown that at $(2,1)$ we have a minimum and so $f(2,1)\leq f(x,y)$ for all $(x,y\in \mathbb{R}^2$.

I did that in this way:
I calculated the gradient and set this equal to zero and found that the only critical point is $(2,1)$. Then I calculated the Hessian matrix, the eigenvalues and since these are positive we get that the critical point is a minimum.
Is this the only way to do that? I mean after having found the critical point could we somehow show directly that $f(2,1)\leq f(x,y)$ for all $(x,y\in \mathbb{R}^2$ without using the Hessian Matrix?

:unsure: Now I want to determine the points of the surface $z^2=8+(x-y)^2$ that have the minimum distance to the point $(3,0,0)$.

I have done the following:

Let $(x,y,z)$ be the point that we are looking for. This point is on the surface, so it satisfies the equation $z^2=8+(x-y)^2$.
The distance to $(3,0,0)$ is $d(x,y,z)=\sqrt{(x-3)^2+y^2+z^2}$.
We can also consider $d^2$ onstead of $d$, or not?
We can substitute $z^2$ by $8+(x-y)^2$, or not?
If yes, then we get $d^2=(x-3)^2+y^2+8+(x-y)^2=(x-3)^2+y^2+(x-y)^2+8=f(x,y)+8$.
So we have the function $f$ plus a constant. Does this mean that the minimum of $d$ is at the same point as the minimum of $f$ ?:unsure:

 

[I like Serena]

Is this the only way to do that? I mean after having found the critical point could we somehow show directly that $f(2,1)\leq f(x,y)$ for all $(x,y\in \mathbb{R}^2$ without using the Hessian Matrix?

Hey mathmari!

The function consists of only squares and it follows that the function value increases to plus infinity in all directions.
In other words, the critical point cannot be a maximum, and it cannot be a saddle point either.
Therefore it must be a minimum.

So we have the function $f$ plus a constant. Does this mean that the minimum of $d$ is at the same point as the minimum of $f$ ?

Yep. (Sun)

 

[mathmari]

The function consists of only squares and it follows that the function value increases to plus infinity in all directions.
In other words, the critical point cannot be a maximum, and it cannot be a saddle point either.
Therefore it must be a minimum.

That it cannot be a maximum is clear to me.
I haven't really understood why it cannot be a saddle point. Could you explain that further to me? :unsure:

 

[I like Serena]

That it cannot be a maximum is clear to me.
I haven't really understood why it cannot be a saddle point. Could you explain that further to me?

If it is a saddle point, then there must be a direction where $f$ goes to up, and also a direction where $f$ goes down.
Since $f$ goes to $+\infty$ in all directions, it's not going down.

 

[mathmari]

If it is a saddle point, then there must be a direction where $f$ goes to up, and also a direction where $f$ goes down.
Since $f$ goes to $+\infty$ in all directions, it's not going down.

I see! Thank you very much!