# Maximum value of f(r)

1. Mar 14, 2009

### trelek2

given:$$f(r)=x ^{2}+3y ^{2} +2z ^{2}$$
The task was to calculate at the point (2,3,1): the grad of f, tangent plane, directional derivative in the direction (2,-1,0) but also to find the maximum value of f subject to the condition that.
$$r ^{2} =1$$
I've done all except the last part, I have no idea what I am supposed to do here, and I don't really understand what they want.

2. Mar 14, 2009

### HallsofIvy

Well, I don't either because there is not "r" given. If I had to guess it would be either $r= x^2+ y^2+ z^2= 1$, although I would be inclined to use "$\rho$", or $r= x^2+ y^2= 1$.

3. Mar 14, 2009

### yyat

Why do you write "f(r)" when f is a function of x,y,z and can not be written as a function of the radius?
To find the maximum you should probably use the Langrange multiplier method (find points where gradient is normal to the set on which f should be optimized).

Last edited by a moderator: May 4, 2017
4. Mar 14, 2009

### trelek2

It is what i have in the exercise...
I also have problems distinguishing r and the x+y+z stuff. How would you treat it?
So supposing r^2=1=x^2+y^2+z^2, should I then take the gradient of f(r) at the given point to find the value of the langrange multiplayer?

Last edited: Mar 14, 2009
5. Mar 14, 2009

### yyat

Apply the Langrange multiplier method (as in the wikipedia article or maybe in your notes/textbook) to the function f(x,y,z)=x^2+3y^2+2z^2 and the constraint g(x,y,z)=x^2+y^2+z^2=1. You have x,y,z instead of just x,y as in the wikipedia article, but you should be able to adapt the formulas easily.

6. Mar 15, 2009

### trelek2

I'm still really confused how to do this. Since I get the gradient of f and the constraint function g in terms of (2x+2xlambda, 6y+2ylambda,4z+2zlambda) It seems to imply that lambda has to be 3 different values at the same time as the variables get reduced. To keep the variables I can take as stated in the exercise at the point f(2,3,1) but does that make any sense?