Proving that a critical point is a minimum

Question

I am solving a problem, which is

"Find the point of the paraboloid $P:=\{(x,y,z)\in\mathbb{R}^3 | x^2+y^2=z\}$ which is the nearest to the point $(1,1,\frac12)$."

I have already determined (using the Lagrange multipliers method) that $p=(2^{-\frac23},2^{-\frac23},2^{-\frac13})$ is a critical point of $f(x,y,z)=(x-1)^2+(y-1)^2+(z-\frac12)^2$ (which is the square of the distance function) with the restraint $g(x,y,z)=0$, where $g(x,y,z)=x^2+y^2-z$.

I need some help in order to prove that $p$ is the minimum of $f|_P$.

If possible, give me only a hint, as it is homework and I only want some idea to finish the problem by myself.

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EDIT: As requested, I'm showing my calculations so far.

A critical point of $f|_P$ is, by the Lagrange multipliers method, one which satisfies $g(x,y,z)=0$ (that is, $x^2+y^2=z$) and $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ for some $\lambda \in \mathbb{R}$.

We have $\nabla f(x,y,z)=(2x-2,2y-2,2z-1)$ and $\nabla g(x,y,z)= (2x,2y,-1)$, which yields the system $$(2x-2=\lambda 2x)\wedge(2y-2=\lambda 2y)\wedge(2z-1=-\lambda),$$ which is equivalent to $$(x=\tfrac{1}{1-\lambda})\wedge (y=\tfrac{1}{1-\lambda}) \wedge(z=\tfrac{1-\lambda}{2}).$$

So we get $x=y$ and $2x^2=z$ (because $x^2+y^2=z$), which gives $\frac{2}{(1-\lambda)^2}=\frac{1-\lambda}{2}$, that is, $\lambda = 1-2^\frac23$.

This gives us $x=y=2^{-\frac23}$ and $z=2^{-\frac13}$.

Answer 1

Probably not the answer you're looking for but...

$z=x^2+y^2$ and $w=(x-1)^2+(y-1)^2+(z-1/2)^2$ where $w$ is the distance to (1,1,1/2) squared. Plugging the first equation into the second yields $w=(x-1)^2+(y-1)^2+(x^2+y^2-1/2)^2 = x^4+2x^2y^2+y^4-2x-2y+9/4$. $w_x=4x^3+4xy^2-2$, $w_y=4x^2y+4y^3-2$, $w_{xx}=12x^2+4y^2$, $w_{xy}=w_{yx}=8xy$, and $w_{yy}=4x^2+12y^2$. The discriminant $D=w_{xx}w_{yy}-(w_{xy})^2$ at the point $(x,y)=(2^{-2/3},2^{-2/3})$ is $24\cdot \sqrt[3]{2}>0$ and $w_{xx}(2^{-2/3},2^{-2/3})=4\cdot 2^{2/3}>0$ so by the second derivative test, $(x,y)=(2^{-2/3},2^{-2/3})$ is a (relative) minimum.

The general problem is that your constraint yields an unbounded set. If the set was compact (closed and bounded), the extreme value theorem would apply and we would know that the global min and max fall among the critical points. But since this set is unbounded the extreme value theorem does not apply and we have to be more careful. Although from an intuitive viewpoint, it seems obvious that this must be a minimum since points on the surface can get arbitrarily far from $(1,1,1/2)$ (so no global max). This really needs to be the minimum.

Addendum (for the unconvinced):

$w(x,y)$ grows as $(x,y)$ tends away from the origin like $x^4+x^2y^2+y^4$ (the lower order terms are irrelevant when $\sqrt{x^2+y^2}$ is large. Thus $w(x,y)$ heads off to $\infty$ as $(x,y)$ move away from the origin.

Pick some very large disk $x^2+y^2 \leq R^2$ ($R \gg 0$). All of the values of $w$ along the boundary of this disk will be larger than the value at our (relative) minimum (since we picked a large $R$ and $w$ is heading off to $\infty$ in all directions). This disk is a compact set, so the extreme value theorem applies. $w$ must attain a global minimum on this set at a critical point or a boundary point. Since there is only one critical point and all of the boundary values are larger than the value at this critical point. It must be the global minimum on this disk. The same argument applies to all larger disks. Thus this is a global min.

Answer 2

Prof. Cook is using the multivariate index $D$ applied to the distance-squared function for points on the paraboloid measured from $(1,1,\frac{1}{2})$. It's not the easiest thing to read, but it does establish that the point you found is at the minimal distance from the point external to the paraboloid.

Alternatively, you could put your results for $x$ and $y$ ($x = y = \frac{1}{2z}$), into the distance squared formula to reduce it to a function of $z$ alone:

$$s^2 = (\frac{1}{2z} - 1)^2 + (\frac{1}{2z} - 1)^2 + (z - \frac{1}{2})^2 = \frac{1}{2z^2} - \frac{2}{z} + 2 + z^2 - z + \frac{1}{4}.$$

The second derivative is $2 - \frac{4}{z^3} + \frac{3}{z^4}$, which is positive for $z = 2^{-1/3}$, so $s^2$ is minimized for that value of $z$ (and thus for the point you found). Since the distance $s$ is non-negative, finding the minimum for $s^2$ suffices.

EDIT: D'oh -- it would have been even easier to use $y=x$ and $z=2x^2$ to write

$$s^2 = (x - 1)^2 + (x - 1)^2 + (2x^2 - \frac{1}{2})^2 = 4x^4 - 4x + \frac{9}{4} \Rightarrow \frac{d^2}{dx^2}(s^2) = 48x^2 , $$which is plainly positive for $x = 2^{-2/3}$; it is also easier to see that $\frac{d}{dx}(s^2) = 16x^3 - 4 = 0$ there. (I'd been working with $z$ as the principal variable the first time through...)

Answer 3

Your paraboloid has a surjective parametrization $$(u,v)\mapsto(u,v,u^2+v^2)\qquad\bigl((u,v)\in{\mathbb R}^2\bigr)\ .$$ Therefore we have to find the minimum of $$f(u,v):=(u-1)^2+(v-1)^2+\left(u^2+v^2-{1\over2}\right)^2\qquad\bigl((u,v)\in{\mathbb R}^2\bigr)\ .$$ As $f$ gets arbitrarily large when $|u|$ or $|v|$ are large it follows that there is a minimal value of $f$ which can be found by solving $\nabla f(u,v)=(0,0)$. The computation gives $$f_u=-2+4u(u^2+v^2)\ ,\quad f_v=-2+4v(u^2+v^2)\ .$$ The condition $f_u=f_v=0$ implies $(u,v)\ne(0,0)$, or $u^2+v^2\ne0$. As a consequence $u=v$ and therefore $0=-2+8u^3$. We conclude that there is a single point $(u,v)$ with $\nabla f(u,v)=(0,0)$, namely the point $\bigl(4^{-1/3}, 4^{-1/3}\bigr)$. The corresponding point on the paraboloid has coordinates $\bigl(4^{-1/3}, 4^{-1/3},2^{-1/3}\bigr)$.