Lagrange Multiplier help

**Rhode963** · March 2nd 2010, 09:57 AM

I've got two Lagrange Multiplier questions I need help on.

They both say "Use Lagrange multipliers to find the maximum and minimum values of the frunction subject to the given constraint(s).

Here is the first:

$f(x,y)=e^{xy}; x^3+y^3=16$
I've gotten to $f_x=(\lambda)g_x$ which is $ye^{xy}=(\lambda)3x^2$ , and $f_y=(\lambda)g_y$ which is $xe^{xy}=(\lambda)(-3y^2)$ and now I'm stuck... I don't know what to do.

The second is $f(x,y,z)=8x-4z; x^2+10y^2+z^2=5$
I've gotten to
$f_x=(\lambda)g_x, --> 8=(\lambda)2x$
$f_y=(\lambda)g_y, --> 0=(\lambda)20y$
$f_z=(\lambda)g_z, --> -4=(\lambda)2z$
And now I'm stuck.

**Aryth** · March 2nd 2010, 01:49 PM

I'll give you the second one... And maybe that'll show you the method and you may be able to solve the first one, if not I'll help you there as well.

Alright, so your given:

$f(x,y,z) = 8x - 4z$

Bound by the constraint function: $\phi(x,y,z) = x^2 + 10y^2 + z^2 = 5$

You've already shown that the components of the gradients are related by the constant multiple $\lambda$ , which gave you:

$8 = 2\lambda x$

$0 = 20\lambda y$

$-4 = 2\lambda z$

First thing is to try to solve each equation for a different variable if possible:

$\frac{4}{\lambda} = x$

$0 = y$ (We can conclude this because neither x nor z can be undefined, and therefore $\lambda \neq 0$ )

$-\frac{2}{\lambda} = z$

You can see right off the bat that x and z are similar, but more specifically are:

$-\frac{2}{\lambda} = z = -\frac{1}{2}\frac{4}{\lambda} = -\frac{1}{2}x$

Now, you plug this all in the constraint function

$x^2 + 10(0)^2 + \left(-\frac{1}{2}x\right)^2 = 5$

$x^2 + \frac{x^2}{4} = 5$

$\frac{5}{4}x^2 = 5$

$x^2 = 2$

$x = \pm 2$

Now you plug this into the formula for z:

$z = -\frac{1}{2}(\pm 2) = \mp 1$

So, the critical points are:

$(2,0,-1)$ and $(-2, 0 , 1)$

You can plug these in on your own to find out whether they are a max or a min.

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