Lagrange Multipliers, help with solving system of equations

**sgcb** · October 12th 2012, 06:42 PM

Use Lagrange multipliers to find the maximum and minumum values of the function subject to the given constraint

$f(x, y, z) = xyz$
$g(x, y, z) = x^2+2y^2+3z^2 = 6$

$\nabla f = [yz, xz, xy]$
$\nabla g = [2x, 4y, 6z]$

$yz - \lambda 2x = 0$
$xz - \lambda 4y = 0$
$xy - \lambda 6z = 0$
$x^2+2y^2+3z^2 = 6$

I've gotten the system of equations. (shown above) I'm just not sure how to go about tackling this one. Any words of wisdom from someone more clever would be appreciated

**MaxJasper** · October 12th 2012, 07:49 PM

Play around a bit with those multipliers to reach:

Max(xyz) = $\frac{2}{\sqrt{3}}$

for: {x,y,z} = $\left\{x\to \sqrt{2},y\to -1,z\to -\sqrt{\frac{2}{3}}\right\}$

**sgcb** · October 12th 2012, 08:01 PM

Can you expand a bit on what you mean by "play around", please? Something more to go on? I've been playing around off and on this afternoon and so far am seeing little progress. That is the answer in back of book though.

**sgcb** · October 12th 2012, 08:55 PM

Ah, here we are:

$\lambda = \frac{yz}{2x}$ so $y^2 = \frac{x^2}{2}$ and $z^2 = \frac{x^2}{3}$

plugging into constraint gives:
$x = \pm \sqrt 2$

so
$f(\sqrt{2}, 1, \sqrt{\frac{2}{3}}) = \frac{2}{\sqrt{3}}$ max
$f(-\sqrt{2}, -1, -\sqrt{\frac{2}{3}}) = -\frac{2}{\sqrt{3}}$ min

**HallsofIvy** · October 13th 2012, 10:19 AM

Since a specific value of $\lambda$ is not necessary for a solution, I find it is often easiest to eliminate $\lambda$ first by dividing one equation by another. Here, I would first rewrite the equations as $yz= \lambda 2x$ , $xz= \lambda 4y$ , and $xy= \lambda 6z$ .

Dividing the first equation by the second, $\frac{y}{x}= \frac{x}{2y}$ which is the same as $2y^2= x^2$ . Similarly, dividing the second equation by the third, $\frac{z}{y}= \frac{2y}{3z}$ which is the same as $3z^2= 2y^2$ .

The constraint that has to be satisied is $x^2+ 2y^2+ 3z^2= 6$ and, from the above, we can replace both $x^2$ and $3z^2$ by $2y^2$ making the equation $6y^2= 6$ so that $y^2= 1$ and $y= \pm 1$ . With $y= \pm 1$ , $x^2= 2y^2= 2$ so that $x= \pm\sqrt{2}$ and $3z^2= 2y^2= 2$ so that $z^2= \frac{2}{3}$ and $z= \pm\sqrt{\frac{2}{3}}= \pm\frac{\sqrt{6}}{3}$ . That gives a total of eight points that should be checked for max and min values.

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