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!
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.