# Maximum value of a multivariable function

• Sep 2nd 2008, 05:28 PM
Paperwings
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 $\mathbb{R}^3$
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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.
• Sep 2nd 2008, 07:47 PM
NonCommAlg
Quote:

Originally Posted by Paperwings
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 $u=, \ v = <1,2,3>.$ 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.