LaGrange Multiplier Problem

May 2010
20
1
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?
\(\displaystyle 4\lambda x^3=4\)
\(\displaystyle 4\lambda y^3=-1/2\)
\(\displaystyle 4\lambda z^3=27/2\)
\(\displaystyle x^4+y^4+z^4=1\)
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
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.
 
May 2010
20
1
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?
 
Last edited:

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
\(\displaystyle 4\lambda x^3=4\implies x^3= \frac{4}{4\lambda }\implies x^3= \frac{1}{\lambda }\implies x= \frac{1}{\sqrt[3]{\lambda}}\)

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 \(\displaystyle x,y,z\)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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.