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Euclidean Space - Maximum Value

  • #1

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.
 

Answers and Replies

  • #2
148
0
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
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:
  • #4
148
0
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
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
148
0
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
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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