Lagrange multipliers - maximum and minimum values given constraint Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer does not exist, enter DNE.)
$ \ \ f(x, y, z) = xyz \ ; \ \    x^2 + 2y^2 + 3z^2 = 96$
What I have gotten to:
$\Delta f = \ <yz, xz, xy>$
and $\Delta g = \ λ<2x, 4y, 6z>$
set them equal and get:
$x^2 = 2y^2$ and 
$z^2 = \frac{2}{3} y^2$ .
Then:
$$
\begin{cases}x^2 + 2y^2 +3z^2 = 96 \\
6y^2 = 96 \\
y = \pm16 \end{cases}
$$
Plugging $y$ into $z^2$ and $x^2$ equations above, you get the points:
$(\pm16\sqrt{2}, \ \pm 16, \  \pm16\sqrt{2/3} ) \ .$
Plugging the points back into $f$, I got a maximum of $32\sqrt{3}$ and a minimum of $-32\sqrt{3}$.
Where did I go wrong?
Thank you in advance!
 A: I do not see where you used Lagrange multipliers. Any critical points will satisfy the Lagrange multipliers equation
$$
\begin{bmatrix}
yz & xz & xy
\end{bmatrix} = \lambda \begin{bmatrix}
2x & 4y & 6z
\end{bmatrix} \ .
$$
This gives you the system of equations
$$
\begin{align}
yz &= 2\lambda x & (1) \\
xz &= 4\lambda y & (2) \\
xy &= 6\lambda z & (3) \\
96 &= x^2+2y^2+3z^2 & (4)
\end{align}
$$
First consider $\lambda = 0$ and then consider other cases. Try this and see how it goes for you. I hope this helps.
EDIT:
If you assume $x,y,z\neq 0$, you can solve for $\lambda$ in each of $(1), (2), (3),$ and $(4)$. You then obtain (by equating these $\lambda$), as you did, $6y^2=96$, which gives $y=\pm4$. You similarly obtain $x^2 = 32$, or $x=\pm4\sqrt{2}$, and $z^2 = \frac{32}{3}$, or $z=\pm4\sqrt{\frac{2}{3}}$. With all of the $\pm$'s, you get a few critical points. Plug them in to see which ones are the largest/smallest.
A: With the the combination of critical points, the global max would be 128/3 and global min would be -128/3. Just any combination of multiple of negatives or positives that produces the positive(max) end and negative end(min).
