# Thread: Max and min values of multivariable functions

1. ## Max and min values of multivariable functions

In general I can figure out max and min values, but I'm have some trouble with the added constraints. Here's the problem:

"Find the maximum and minimum values of the function f(x,y)=x^2+y^2+z^2 subject to the constraint x^4+y^4+z^4=1"

2. ## Re: Max and min values of multivariable functions

Using the Lagrange multiplier method you can form an auxiliary function

$\displaystyle L(x,y,z,\lambda) = (x^2+y^2+z^2)+\lambda (x^4+y^4+z^4-1)$

The stationary points can then be found by solving $\displaystyle \nabla L = 0$

3. ## Re: Max and min values of multivariable functions

Do i include the partial derivative of gamma in the Gradient to find the points, or do i only use x, y, and z for the gradient still? Sorry we haven't covered much Lagrange yet. Just starting it actually.

4. ## Re: Max and min values of multivariable functions

Get them all. Why would we go to the trouble to define an extra variable and then just ignore it?

5. ## Re: Max and min values of multivariable functions

I think you mean lambda, yes take the derivative with respect to lambda and x,y,z.

You only need the result for x,y,z in context of the original function.

6. ## Re: Max and min values of multivariable functions

Yeah i meant lambda. I mix up greek letters quite a bit.

I didn't mean do I ignore it, I've just never seen a gradient include lambda before and I wasn't sure how to include it.

7. ## Re: Max and min values of multivariable functions

Once you start to studying the method it will become more clear why its included.

Good luck!

8. ## Re: Max and min values of multivariable functions

Okay so tell me if I get this correct or not. Pardon the lack of notation, I don't know how to type them.

2x+4(lambda)x^3=0
2y+4(lambda)y^3=0
2z+4(lambda)z^3=0
x^4+y^4+z^4-1=0

x=sqrt(-1/(2*lambda))
y=sqrt(-1/(2*lambda))
z=sqrt(-1/(2*lambda))

(sqrt(-1/(2*lambda)))^4+(sqrt(-1/(2*lambda)))^4+(sqrt(-1/(2*lambda)))^4-1=0

lambda=sqrt(3)/2

And then i plug lambda back into the first three equations that are equal to 0.