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Homework Help: Euclidean Space - Maximum Value

  1. Jul 4, 2008 #1
    1. The problem statement, all variables and given/known data
    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]

    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 [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.
     
  2. jcsd
  3. Jul 4, 2008 #2
    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?
     
  4. Jul 4, 2008 #3
    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: Jul 4, 2008
  5. Jul 4, 2008 #4
    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?
     
  6. Jul 4, 2008 #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?
     
  7. Jul 4, 2008 #6
    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?
     
  8. Jul 4, 2008 #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
    Thank you very much for your help. Cheers.
     
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