# Thread: Lagrange Multipliers, help with solving system of equations

1. ## Lagrange Multipliers, help with solving system of equations

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

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

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

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

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

2. ## Re: Lagrange Multipliers, help with solving system of equations

Play around a bit with those multipliers to reach:

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

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

3. ## Re: Lagrange Multipliers, help with solving system of equations

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.

4. ## Re: Lagrange Multipliers, help with solving system of equations

Ah, here we are:

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

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

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

5. ## Re: Lagrange Multipliers, help with solving system of equations

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

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

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

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