Euclidean Space - Maximum Value

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Homework Help Overview

The problem involves finding the maximum value of the expression \(\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}\) as the vector \((x,y,z)\) varies among nonzero points in \(\mathbb{R}^3\). The context relates to scalar products, norms, and the geometry of vectors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find extreme values by calculating partial derivatives and setting up a system of equations but expresses uncertainty about the next steps.
  • Some participants suggest considering the dot product of vectors and the implications of directional derivatives, questioning the necessity of differentiation.
  • Others explore the geometric interpretation of the dot product and its maximum value based on the direction of the vectors involved.

Discussion Status

The discussion is active with various approaches being explored, including algebraic manipulation and geometric interpretations. Some participants have provided insights that may guide the original poster towards a clearer understanding of the problem, although no consensus has been reached.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the requirement that \((x,y,z)\) must be nonzero, which influences their reasoning about the maximum value of the expression.

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Homework Statement


Find the maximum of
[tex]\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}[/tex]
as (x,y,z) varies among nonzero points in [tex]R^{3}[/tex]

Homework 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.


The Attempt at a Solution


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

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

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

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

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 [tex]\left\| a \right\| = \sqrt{1^2+2^2+3^2} = \sqrt{14}[/tex]
2. So, the unit vector u would be
[tex]u=\frac{1}{ \sqrt{14}}a = \frac{1}{ \sqrt{14}}i + \frac{2}{ \sqrt{14}}j + \frac{3}{ \sqrt{14}}k[/tex]
3. Thus, the directional derivative at (1,2,3) is:
[tex]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}}[/tex]
4. Last, plug in (1,2,3) in the partial deratives to obtain the maximum value?
 
Last edited:
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?
 
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?
 
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?
 
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| = [tex]\sqrt{1^2+2^2+3^2} = \sqrt{14}[/tex]

I finally understood what you meant by
Notice that this thing is the dot product of (1,2,3) and the unit vector in the direction of (x,y,z).

Thank you very much for your help. Cheers.
 

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