# LaGrange Multiplier Problem

#### 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?
$$\displaystyle 4\lambda x^3=4$$
$$\displaystyle 4\lambda y^3=-1/2$$
$$\displaystyle 4\lambda z^3=27/2$$
$$\displaystyle x^4+y^4+z^4=1$$

#### pickslides

MHF Helper
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.

#### 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?

Last edited:

#### pickslides

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

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
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.