Use Lagrange multipliers to find minimum and maximum

Question

$$f(x,y,z) = x^{2}y^{2}z^{2}$$ If: $$g(x,y,z)=x^{2}+y^{2}+z^{2}+1 = 0$$

The method I know is to create the following function:

$F(x,y,z,\lambda)=f(x,y,z)-\lambda g(x,y,z)$

Then create system of equations:

$\frac{dF}{dx}=0$
$\frac{dF}{dy}=0$
$\frac{dF}{dz}=0$
$\frac{dF}{d\lambda}=0$

The results are possible minimum/maximum points.

Then I need to create Hessian matrix with 0 as top left value. If the determinant is positive, it's maximum, etc...

My problem is - It quite doesn't work here, and I don't know why. The system of equation gives $x^{2}=y^{2}=z^{2}$, so saying that $r=x^{2}=y^{2}=z^{2}$ it's $P=(r,r,r)$.

The Hessian matrix gets really complicated.

$$ H = \left[ \begin{array}{ccc} 0 & 2x & 2y & 2z \\ 2x & 2y^{2}z^{2} - 2\lambda & 4zxy^{2} & 4zxy^{2} \\ 2y & 4xyz^{2} & 2x^{2}z^{2} - 2\lambda & 4zx^{2}y \\ 2z & 4xy^{2}z & 4x^{2}y^{2} & 2x^{2}y^{2} - 2\lambda \end{array} \right] $$

And I'm stuck here, what's wrong?

I don't understand: why do you say the Hessian's left top entry has to be zero? — DonAntonio, Aug 02 '13 at 11:14
Hm ... there are no points $(x,y,z) \in \mathbb R^3$ with $g(x,y,z) = x^2 + y^2 + z^2 + 1 = 0$, as $g \ge 1$. — martini, Aug 02 '13 at 11:15
Hmmm, you're right...I must have mistaken the sign before "1", so it should be $x^{2}+y^{2}+z^{2}-1=0$ — khernik, Aug 02 '13 at 11:17
@DanielFischer I was also irrated at first, note that $\frac{\partial F}{\partial \lambda} = g$, so this line says $g(x,y,z) = 0$ and belongs there. — martini, Aug 02 '13 at 11:28
@martini Ah, I see. We never formulated the constraints in that way, we always just said $g = 0$. — Daniel Fischer, Aug 02 '13 at 11:33
if $x^{2}+y^{2}+z^{2}-1=0$, with AM-GM, you can get MAX at once. min=0 is trivial. — chenbai, Aug 04 '13 at 03:34

score 1 · Accepted Answer · answered Aug 02 '13 at 11:33

Your equations read \begin{align*} 2xy^2z^2 - 2\lambda x &= 0\\ 2x^2yz^2 - 2\lambda y &= 0\\ 2x^2y^2z - 2\lambda z &= 0\\ x^2 + y^2 + z^2 - 1 &= 0 \end{align*} We either have $\lambda = 0$, then one of $x, y, z$ has to be zero also, giving three families of solutions, in the case $\lambda \ne 0$, we have $xyz \ne 0$, giving $$ \lambda = y^2z^2 = x^2z^2 = y^2z^2 $$ so $x^2 = y^2 = z^2 = \frac 13$ (as $x^2 + y^2 + z^2 - 1 = 0$). Now plug in all solutions $(\pm \frac 1{\sqrt 3}, \pm \frac 1{\sqrt 3}, \pm \frac 1{\sqrt 3})$ and the ones found above into the Hessian, if you want, but you can also note that $f$ is 0 on the $\lambda =0$-type solutions and $f = \frac 1{27}$ on the second type. So $f$'s maximum is $\frac 1{27}$, as $f$ must attain its maximum on the compact $S^2$.

score 0 · Answer 2 · answered Aug 02 '13 at 11:34

I get:

$$\begin{align*}F'_x&=2xy^2z^2-2\lambda x=0\\ F'_y&=2x^2yz^2-2\lambda y=0\\ F'_z&=2x^2y^2z-2\lambda z=0\\ F'_\lambda&=x^2+y^2+z^2-1=0\end{align*}$$

Solving for $\,\lambda\;$ and equating we get (assuming $\;x,y,z\ne 0\;$)

$$\frac1{y^2z^2}=\frac1{x^2z^2}=\frac1{x^2y^2}\implies \pm x=\pm y=\pm z$$

and from here, puting $\,x=y=z\;$ (or their additive inverses):

$$3x^2=1\implies x=\pm\frac1{\sqrt3}$$

and now form the Hessian...

Use Lagrange multipliers to find minimum and maximum

2 Answers2