maximum value of the expression : $2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1$

Question

I have to find out the maximum value of the expression :

$2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1.$

Answer 1

By Cauchy Schwarz inequality $$(2x+3y+z)^2\le (4+9+1)(x^2+y^2+z^2)$$

$\therefore$ the maxium value that can be attained is when $2x=3y=z$ , and the value attained is $\sqrt{14}$

Answer 2

When I see $f(x,y,z) = 2x + 3y+ z$, I think of a collection of parallel planes. As the distance from the origin increases f(x,y) increases. So what is the plane that is tangent to the sphere?

Follow the normal vector (2,3,1) until it intersects the surface of the sphere.

Answer 3

Let $(x,y,z)=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) \in S^2$

\begin{align*} 2\cos \phi+3\sin \phi & \le \sqrt{2^2+3^2} \\ 2x+3y &= \sin \theta (2\cos \phi+3\sin \phi) \\ & \le \sqrt{13} \, |\sin \theta| \\ 2x+3y+z & \le \sqrt{13} \, |\sin \theta|+\cos \theta \\ & \le \sqrt{13+1} \\ &= \sqrt{14} \end{align*}