Sufficient criteria for specific constrained extrema

Question

I'm trying to solve the next problem using lagrange multipliers:
Find the distance from the origin to the curve $z^2=x^2+y^2$, $x-2z = 3$.
We define the Lagrangian $$ \mathcal{L}(x,y,z,\lambda,\mu) = x^2+y^2+z^2 + \lambda(x^2+y^2-z^2)+\mu(x-2z-3) $$ and found two solutions $(1,0,-1,-\frac{1}{3},-\frac{4}{3})$ and $(-3,0,-3,-3,-12)$. So the two candidates for extrema are $(1,0,-1)$ and $(-3,0,-3)$. Graphically it is obvious that they are minimum and maximum respectively. Now the problem arises when we try to prove it using some criteria. I tried using the Hessian of the distance square function, but it's not useful because the candidates are not critical point of the distance square. The other possible method that I'm aware at my level of math is to study the bordered hessian but one of the determinant for both of the point is zero, so the criteria is not useful. Is there some criteria (not too advanced) that could be useful for this problem?

Answer 1

The surface $z^2 = x^2 + y^2$ is a cone that forms both above $z = 0$ and below it, and z-axis is the axis of cone. The plane $x-2z = 3$ is parallel to y-axis. It is easy to visualize where the maximum and minimum distance to origin occur. Here is a simple approach -

The curve is given by intersection of $z^2=x^2+y^2, x - 2z = 3$. The projection in XY-plane is,

$(x-3)^2 = 4z^2 = 4 (x^2+y^2) \implies 3x^2 + 6x + 4y^2 = 9$

So the intersection curve is, $\dfrac{(x+1)^2}{4} + \dfrac{y^2}{3} = 1 \ $ and $ \ z = \dfrac{x-3}{2}$

Parametrize the curve as $ \ r(\theta)= (-1 + 2 \cos\theta, \sqrt3 \sin\theta, -2 + \cos\theta), \ 0 \leq \theta \leq 2\pi$.

So if distance from origin to a point on the curve is $d$,

$d^2 = x^2 + y^2 + z^2 = (-1 + 2 \cos\theta)^2 + 3 \sin^2\theta + (-2 + \cos\theta)^2$

$ = 8 + 2 \cos^2\theta - 8 \cos\theta = 2 \cdot (2 - \cos\theta)^2$

So we can clearly see that $|d|$ is maximum when $\cos\theta = -1 \implies \theta = \pi$ and it is minimum when $\cos\theta = 1 \implies \theta = 0 \ (\text{ or } 2\pi)$

That leads to maximum distance of $3 \sqrt2$ and minimum distance of $\sqrt2$. You can also plug in value of $\theta$ in $r(\theta)$ to find points where maximum and minimum distance from origin occur.