# Find the Lagrange multipliers with one constraint: $f(x,y,z) = xyz$ and $g(x,y,z) = x^2+2y^2+3z^2 = 6$

Where $f(x,y,z) = xyz$ and the constraint is $g(x,y,z) = x^2+2y^2+3z^2 = 6$

I have tried this problem like three or four times and not gotten the solution, I even asked this question once and got the wrong answer from my one and only answerer. The correct answer is {$\pm\sqrt{2}$,$\pm1$,$\pm\frac{\sqrt{2}}{\sqrt{3}}$} This problem just makes no sense from an algebraic standpoint this is my 4th problem off the odd number exercises and I could do the previous 3 just fine.

My attempt:

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

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

$$F=x y z +\lambda \left(x^2+2 y^2+3 z^2-6\right)$$ Computing derivatives $$F'_x=y z+2 \lambda x=0\tag 1$$ $$F'_y=x z+4 \lambda y=0\tag 2$$ $$F'_z=x y+6 \lambda z=0\tag 3$$ $$F'_\lambda=x^2+2 y^2+3 z^2-6=0\tag 4$$ 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 $4y^2=6z^2$ and got $\frac{6}{4}$ = $\frac{3}{2}$ Now plugging these into equation 4 I obtained $3+\frac32+3z^2-6=0$ 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 $F$.

Multiplying equations 1,2,3 by $x,y,z$ we obtain that $2x^2=4y^2=6z^2$.

This is correct. (And, since you know what the solution should be, you can check for yourself, that the solution fulfills this equation.)

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 $4y^2=6z^2$ and got $\frac{6}{4}$ = $\frac{3}{2}$

This is not true.

From $2x^2=6z^2$ you get $x^2=3z^2$.

From $4y^2=6z^2$ you get $2y^2=3z^2$ or $y^2=\frac32z^2$.

Now by plugging this into the equation $x^2+2y^2+3z^2=6$ you get $3z^2+3z^2+3z^2=6$, i.e. $$9z^2=6$$ and $z^2=\frac23$, $z=\pm\sqrt{\frac23}$.