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I have $f(x,y, z)=xyz$ with constraint $x^2+2y^2+3z^2=6$

I know $f(x,y, z)=\lambda g(x,y)$

$f_x=\lambda g_x$ , $f_y=\lambda g_y$

Tried

$yz=\lambda 2x$ , $xz=\lambda 4y$ , $xy=6z\lambda$

I first tried solving for lambda in one equation and then plugging that into the other two, but just got $x=\sqrt{8}=z$, $y=4$ which doesn't work in the constraint. I also tried something like what I seen in another problem when they took, for example the $yz=2x\lambda$ and multiplied both sides by the missing variable, this case x, so all the equations had xyz on the left hand side and tried to solve for them being equal and working in the constrain equation getting $x=\sqrt{6}$, $y=\sqrt{3}$ , $z=\sqrt{2}$, in each case the other two variables being $0$. But that just gives zeros for all when put into f(x,y,z). The answer is supposed to be min $\frac{-2}{\sqrt{3}}$ and max at $\frac{2}{\sqrt{3}}$

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    $\begingroup$ What is $z$? is it a constant? I mean whay $f$ is only a function of $x,y$? $\endgroup$ – polfosol Oct 19 '16 at 5:16
  • $\begingroup$ Oh sorry, it's jut suck a habit just doing f(x,y) after so many problems that I just typed it automatically. $\endgroup$ – windy401 Oct 19 '16 at 5:17
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First, there is a solution since the feasible set is compact.

Lagrange multipliers gives $yz + 2 \lambda x = 0$, $xz + 4 \lambda y = 0$, $xy + 6 \lambda z = 0$.

Multiply these equations by $x,y,z$ respectively to get $xyz + 2 \lambda x^2 = 0$, $xyz + 4 \lambda y^2 = 0$, $xyz + 6 \lambda z^2 = 0$, from which we get $\lambda x^2 = 2 \lambda y^2 = 3 \lambda z^2$.

If $\lambda = 0$, then since $xyz=0$, this means one of $x,y,z$ must be zero and it is easy to check that this is not optimal (for example, look at the cost at the point $\pm(\sqrt{2}, 1, \sqrt{2 \over 3})$), hence $\lambda \neq 0$. Dividing through gives $x^2 = 2 y^2 = 3 z^2$.

From the constraint, we see that $x^2 = 2, 2 y^2 = 2, 3 z^2 = 2$. The answers follow from this.

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  • $\begingroup$ Thanks. Good to see I wasn't to far off track. Could you go into more detail about how $\lambda = 0$ was ruled out though? $\endgroup$ – windy401 Oct 19 '16 at 5:58
  • $\begingroup$ If $\lambda = 0$ then two of $x,y,z$ must be zero (fiddle with the equations). Compare the answer obtained here to those and pick a max. $\endgroup$ – GiantTortoise1729 Oct 19 '16 at 6:04
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This proof uses only AM-GM inequality and not Lagrange multipliers. \begin{align*} 2 = \frac{x^2+2y^2+3z^2}{3} \geq (x^2\cdot 2y^2 \cdot 3z^2)^{1/3} = (6x^2y^2z^2)^{1/3} \end{align*} Hence \begin{align*} x^2y^2z^2 \leq \frac{8}{6} \end{align*} and hence \begin{align*} |xyz| \leq \frac{2}{\sqrt{3}} \end{align*}

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