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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!

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    $\begingroup$ your edit is strange, we write $\nabla f=\lambda\nabla g$. $\endgroup$ – Brady Trainor Mar 28 '13 at 0:32
  • $\begingroup$ That's what I meant. Can you help me find where I went wrong? $\endgroup$ – Jason Mar 28 '13 at 0:39
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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.

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    $\begingroup$ assume all variables are non-zero, since otherwise f=0, and we ignore that case for the time-being. $\endgroup$ – Brady Trainor Mar 28 '13 at 0:25
  • $\begingroup$ I did use lambda. Someone edited it out by mistake. $\endgroup$ – Jason Mar 28 '13 at 0:30
  • $\begingroup$ I still need help. $\endgroup$ – Jason Mar 28 '13 at 0:31
  • $\begingroup$ @Ryan: I added some more to my answer. Hopefully this will help (and my algebra is not wrong). I did not check the critical points from the other cases, so you may have other cases to explore. $\endgroup$ – user59083 Mar 28 '13 at 1:00
  • $\begingroup$ Wouldn't 6y^2 = 96, give you y = +-4? $\endgroup$ – Jason Mar 28 '13 at 1:03
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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).

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