Find such points on the ellipsoid $x^2+2y^2+4z^2 = 8$ that are the farthest and nearest to the point $(0;0;3)$

Question

So, I need to compute this optimisation exercise using Lagrange multiplier. I know how to do these sorts of exercises when a constraint function is given, but in this case there's only a point. So, I don't know how to interpret it. Is it meant to be that there is three constraint functions $x=0, y=0 \text{ and } z=0 \text{ or how to interpret it?}$

Answer 1

The way to interpret this problem is to find extrema of the function $$f(x, y, z) = x^2 + y^2 + (z - 3)^2$$ which represents the square distance between a point $(x, y, z)$ and the point $(0, 0, 3)$. Here you are subject to the constraint $$ g(x, y, z) = \frac{x^2}{8} + \frac{y^2}{4} + \frac{z^2}{2} - 1 = 0$$ which is the set of points on your ellipse (I've just divided by $8$ and subtracted $1$ on both sides).

The method of Lagrange multipliers dictates that in a problem with one constraint $g = 0$, an extremum of $f$ is found when $$\nabla f = \lambda \nabla g$$ Plugging in the $f$ and $g$ we have, we obtain the following systems of equations: $$\begin{cases} 2x = \lambda x/4 \\ 2y = \lambda y / 2 \\ 2z - 6 = \lambda z \\ g = 0\end{cases}$$

There a few different cases we must consider here:

• The first is when $\lambda = 0$, in which case $z = 3$ (from the third equation). However, no point on the constraint has $z = 3$, so we can eliminate this case.

Note that if $\lambda \neq 0$, then at least one of $x$ or $y$ must be zero (otherwise we would have $\lambda = 8$ from the first equation and $\lambda = 4$ from the second).

• If $x = 0$ and $y \neq 0$, then $\lambda = 4$, forcing $z = -3$ from the third equation. But no point on the constraint has $z = -3$, so we can rule out this case.

• If $y = 0$ and $x \neq 0$, then $\lambda = 8$, forcing $z = -1$. From the last equation, this would force $x^2 = 4 \iff x = \pm 2$, making the solutions in this case to be the points $(2, 0, -1)$ and $(-2, 0, -1)$.

• Finally, if both $x$ and $y$ are zero, then the fourth equation forces $z^2 = 2 \iff z = \pm \sqrt2$, with $\lambda = 2 \mp 3 \sqrt{2}$. So this case yields the points $(0, 0, \sqrt{2})$ and $(0,0, -\sqrt 2)$.

The only Lagrange points left standing are the points $(\pm 2, 0, -1)$ and $(0, 0, \pm \sqrt{2})$. What remains is to compare the values of $f$ at each of these four points to determine which is a global minimizer/maximizer. Can you proceed from here?