# 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 $F$.

• canh you post your equation system please? May 14, 2016 at 5:28
• Correction to algebra posted as comment to answer in you original post .. It really is incumbent on you to check the algebra in a proposed answer. May 14, 2016 at 5:42
• Possible duplicate of I need help finding the maximum and minimum values according a given constraint May 14, 2016 at 5:44
• Another duplicate: math.stackexchange.com/questions/1277607/… May 14, 2016 at 6:59
• After your question was put on hold, you should not have posted it again. Instead of that you should have added some context. In this case, since you want to know where you have made a mistake, the best way is to show what have you tried so far. I have copied your attempt from the other question you posted. And I have added (solution-verification) tag. Let's wait and see if this is sufficient for reopening the question. May 19, 2016 at 6:02

You asked where exactly you have made a mistake:

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 $z^2=\frac23$, $z=\pm\sqrt{\frac23}$.