# Euclidean Space - Maximum Value

1. Jul 4, 2008

### Paper Wings

1. The problem statement, all variables and given/known data
Find the maximum of
$$\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
as (x,y,z) varies among nonzero points in $$R^{3}$$

2. Relevant equations
I'm not sure. The section in which this problem lies in talks about scalar products, norms, distances of vectors, and orthognality. However, I fail to see how that helps to find the maximum value.

3. The attempt at a solution
1. I set $$f(x,y,z) = \frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$ and find the extreme values at(x,y,z) where $$x \neq 0, y \neq 0, z \neq 0$$ since the problem states that it varies among nonzero points.

2. I find the parital derivaties at (x,y,z).
denom = $$(x^2+y^2+z^2)^{3/2}$$
a. $$f_{x}(x,y,z)= \frac{y^2-2xy+z^2-3xz}{denom}$$
b. $$f_{y}(x,y,z)= \frac{2x^2-yx+2z^2-3yz}{denom}$$
c. $$f_{z}(x,y,z)= \frac{3x^2-xz+3y^2-2yz}{denom}$$

3. I set up a systems of equations
a. $$3x^2-zx+3y^2-2yz=0$$
b. $$2x^2-yx+2z^2-3yz=0$$
c. $$y^2-2xy+z^2-3xz = 0$$

At this point, I'm stuck. Is there a different approach or am I completely off track? Thank you.

2. Jul 4, 2008

### durt

Notice that this thing is the dot product of (1,2,3) and the unit vector in the direction of (x,y,z). Which direction maximizes this?

3. Jul 4, 2008

### Paper Wings

Ah, ok. So, I'm supposed to find the directional derivative of f at (1,2,3) in the direction of a = (x,y,z)

1. Let a be a nonzero vector, where I am trying to find the directional derivatie of f at (x,y,z) in the direction of a $$\left\| a \right\| = \sqrt{1^2+2^2+3^2} = \sqrt{14}$$
2. So, the unit vector u would be
$$u=\frac{1}{ \sqrt{14}}a = \frac{1}{ \sqrt{14}}i + \frac{2}{ \sqrt{14}}j + \frac{3}{ \sqrt{14}}k$$
3. Thus, the directional derivative at (1,2,3) is:
$$D_{u}f(1,2,3)=f_{x}(1,2,3) \frac{1}{ \sqrt{14}} +f_{y}(1,2,3) \frac{2}{ \sqrt{14}}+ f_{z}(1,2,3) \frac{3}{ \sqrt{14}}$$
4. Last, plug in (1,2,3) in the partial deratives to obtain the maximum value?

Last edited: Jul 4, 2008
4. Jul 4, 2008

### durt

Hmmm you're still over complicating things. You don't need any differentiation or anything. If you take the dot product of a vector and another vector of magnitude 1, what is the maximum value of this dot product?

5. Jul 4, 2008

### Paper Wings

Hm, ok. So if the magnitude of the vector is 1, then the maximum value of this dot product would be 1(1) + 2(1) + 3(1) = 6 since the points are nonzero, correct?

6. Jul 4, 2008

### durt

No, (1,1,1) has magnitude sqrt(3). Try to think of this geometrically. Hint: a dot b = |a| |b| cos(theta). What happens if |b| is fixed at 1?

7. Jul 4, 2008

### Paper Wings

If |b| is fixed at 1, then a dot b = |a| cos(theta).

cos(theta) = 1, since a is in the direction of b. So, we get

a dot b = |a| = $$\sqrt{1^2+2^2+3^2} = \sqrt{14}$$

I finally understood what you meant by
Thank you very much for your help. Cheers.