# LaGrange Multiplier Problem

• May 26th 2010, 04:55 PM
keysar7
LaGrange Multiplier Problem
I'm trying to find the minimum and maximum values of
$f(x,y,z,) = (4x - 1/2y + 27/2 z)$

on the surface of $g=x^4+y^4+z^4=1$

I have already put $\bigtriangledown f = \lambda \bigtriangledown g$
Getting the following 4 equations... Where do I go from here?
$4\lambda x^3=4$
$4\lambda y^3=-1/2$
$4\lambda z^3=27/2$
$x^4+y^4+z^4=1$
• May 26th 2010, 05:15 PM
pickslides
Quote:

Originally Posted by keysar7
Getting the following 4 equations... Where do I go from here?
$4\lambda x^3=4$
$4\lambda y^3=-1/2$
$4\lambda z^3=27/2$
$x^4+y^4+z^4=1$

Solve the system, you have equations and 4 unknowns.
• May 26th 2010, 05:19 PM
keysar7
I know I need to solve...
There are 4 unknowns (counting λ) but variables x, y, z are raised to the power of 4.

I don't know how to solve this system... Any help?
• May 26th 2010, 05:33 PM
pickslides
$4\lambda x^3=4\implies x^3= \frac{4}{4\lambda }\implies x^3= \frac{1}{\lambda }\implies x= \frac{1}{\sqrt[3]{\lambda}}$

Do this for the next 2 equations

Then $x^4+y^4+z^4=1$ becomes $\left( \frac{1}{\sqrt[3]{\lambda}}\right)^4+y^4+z^4=1$

Your work will replace $y$ and $z$ with a function of $\lambda$

You will then have a solution for $\lambda$ and use this to find $x,y,z$
• May 27th 2010, 02:17 AM
HallsofIvy
Quote:

Originally Posted by keysar7
I'm trying to find the minimum and maximum values of
$f(x,y,z,) = (4x - 1/2y + 27/2 z)$

on the surface of $g=x^4+y^4+z^4=1$

I have already put $\bigtriangledown f = \lambda \bigtriangledown g$
Getting the following 4 equations... Where do I go from here?
$4\lambda x^3=4$
$4\lambda y^3=-1/2$
$4\lambda z^3=27/2$
$x^4+y^4+z^4=1$

Since the value of $\lambda$ is not part of the solution, I often find it best to eliminate $\lambda$ first by dividing one equation by another.

Dividing the first equation by the second gives $\frac{4\lambda x^3}{4\lambda y^3}= \frac{4}{-1/2}$ or $\frac{x^3}{y^3}= -8$ so $x= -2y$.

Dividing the third equation by the second gives $\frac{4\lambda z^3}{4\lambda y^3}= \frac{27/2}{-1/2}= \frac{z^3}{y^3}= -27$ so $z= -3y$.

Putting x= -2y and z= -3y into the last equation gives a single equation for y.