- #1

mathmari

Gold Member

MHB

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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:

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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