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Math Help - Maximum value of a multivariable function

  1. #1
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    Maximum value of a multivariable function

    Problem:
    Find the maximum value of \frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}} as  (x , y, z) are nonzero points in \mathbb{R}^3
    ========================

    I believe the maximum value of \frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}} is the dot product of the vector a = (1, 2, 3) and the unit vector, b, in the direction of (x, y, z). So  \left\langle a, b \right\rangle = \sqrt{1^2+2^2+3^2} = \sqrt{14} . Does this seem logical? Thank you for your time.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    Problem:
    Find the maximum value of \frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}} as  (x , y, z) are nonzero points in \mathbb{R}^3
    let u=<x,y,z>, \ v = <1,2,3>. then: ||u||=\sqrt{x^2+y^2+z^2}, \ ||v||=\sqrt{14}, and: u \cdot v = x+2y+3z. now by Cauchy Schwartz: |u \cdot v| \leq ||u|| \cdot ||v||.

    thus: |x+2y+3z| \leq \sqrt{14} \sqrt{x^2+y^2+z^2}, which gives us: -\sqrt{14} \leq \frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}} \leq \sqrt{14}. so the maximum and the minimum of your function are \sqrt{14}

    and -\sqrt{14} respectively. Q.E.D.
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