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

• MHB
• mathmari
In summary, the function has a minimum at the point $(2,1)$ and it is not a maximum or a saddle point.
mathmari
Gold Member
MHB
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:

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mathmari said:
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.

mathmari said:
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)

Klaas van Aarsen said:
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:

mathmari said:
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.

Klaas van Aarsen said:
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!

## 1. What does it mean for a point to have a minimum distance to the point (3,0,0)?

Having a minimum distance to the point (3,0,0) means that the given point is the closest point on the surface to the point (3,0,0). This distance is the shortest possible distance between the two points.

## 2. How is the minimum distance to the point (3,0,0) calculated?

The minimum distance to the point (3,0,0) is calculated using the distance formula, which is the square root of the sum of the squared differences between the coordinates of the two points. In this case, the coordinates of the given point on the surface would be substituted for one set of coordinates in the formula, while the coordinates of (3,0,0) would be substituted for the other set.

## 3. Can there be more than one point on the surface with the minimum distance to the point (3,0,0)?

Yes, it is possible for there to be more than one point on the surface with the minimum distance to the point (3,0,0). This would occur if there are multiple points on the surface that are equidistant from the point (3,0,0).

## 4. Are there any practical applications for finding points on a surface with minimum distance to a given point?

Yes, there are many practical applications for this concept. For example, it can be used in navigation systems to find the shortest distance between two points, or in computer graphics to calculate the closest point on a surface for visual effects.

## 5. Can the concept of minimum distance to a point be extended to higher dimensions?

Yes, the concept of minimum distance to a point can be extended to higher dimensions. The distance formula can be used to calculate the minimum distance between points in any number of dimensions, not just three. This concept is important in fields such as physics and mathematics.

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