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I apologize I have asked this question before but it died and I just got around to working it out based on the suggestions so here it is.

Let the function $f$ be defined as $f$($x$,$y$,$z$) $=$ $x$$y$$z$

find the maximum and minimum values subject to the constraint: $g$($x$,$y$,$z$) $=$ $x^2$+2$y^2$+3$z^2$

$$F=x y z +\lambda \left(x^2+2 y^2+3 z^2-a\right)$$ Computing derivatives $$F'_x=y z+2 \lambda x=0\tag 1$$ $$F'_y=x z+4 \lambda y=0\tag 2$$ $$F'_z=x y+6 \lambda z=0\tag 3$$ $$F'_\lambda=x^2+2 y^2+3 z^2-a=0\tag 4$$ Now, I should consider equations $(1,2,3)$ and solve them for $x,y,z$ in terms of $\lambda$.

Multiplying equations 1,2,3 by $x,y,z$ we obtain that $2x^2=4y^2=6z^2$. From here I found the corresponding multiples that $x^2$ and $y^2$ are in terms of $z$ and plugged into equation 4 to solve for z. I found that $x$ Was a multiple of $z$ by 3 and $y$ was a multiple of $z$ by $\frac{3}{2}$ I found this by setting $4y^2$=6$z^2$ and got $\frac{6}{4}$ = $\frac{3}{2}$ Now plugging these into equation 4 I obtained 3+$\frac{3}{2}$+$3z^2-6$=0 My algebra lead me to $z=$ $\pm\frac{1}{\sqrt{2}}$ but if done right z should equal $\pm\frac{\sqrt{2}}{\sqrt{3}}$

I got this by adding 6 over to the right then subtracted 3 leaving me: $\frac{3}{2}$+$3z^2$=$3$ then subtracting $\frac{3}{2}$ lead me to : $3z^2$= $\frac{3}{2}$ and dividing by 3 gives $\frac{3}{2} \div \frac{3}{1}$ which is equivalent to $\frac{3}{2} \times \frac{1}{3}$ = $\frac{3}{6}$ and you can see that really leaves me with $z^2$= $\frac{1}{3}$ which is equivalent to $z=$ $\pm$ $\frac{1}{\sqrt{3}}$

What did I do incorrectly and how do I proceed from here once I have found z. Also how is it that you can write the two functions as two functions added together giving you $F$.

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marked as duplicate by Slade, Did, user147263, Shailesh, colormegone May 19 '16 at 3:49

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