1. ## LaGrange Multiplier Problem

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

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

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

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

3. ## 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?

4. $\displaystyle 4\lambda x^3=4\implies x^3= \frac{4}{4\lambda }\implies x^3= \frac{1}{\lambda }\implies x= \frac{1}{\sqrt[3]{\lambda}}$

Do this for the next 2 equations

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

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

You will then have a solution for $\displaystyle \lambda$ and use this to find $\displaystyle x,y,z$

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

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

I have already put $\displaystyle \bigtriangledown f = \lambda \bigtriangledown g$
Getting the following 4 equations... Where do I go from here?
$\displaystyle 4\lambda x^3=4$
$\displaystyle 4\lambda y^3=-1/2$
$\displaystyle 4\lambda z^3=27/2$
$\displaystyle x^4+y^4+z^4=1$
Since the value of $\displaystyle \lambda$ is not part of the solution, I often find it best to eliminate $\displaystyle \lambda$ first by dividing one equation by another.

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

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

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