<s>[INST] Lagrange Multiplier in 3D Find the minimum and maximum values of the function $f(x,y,z) = x+2y+3z$ where $(x,y,z)$ is on the sphere $x^2+y^2+z^2=1$ using Lagrange multiplier.
So I put them into the Lagrange form and got 
$L(x,y,z,\lambda) = x+2y+3z+ \lambda(x^2+y^2+z^2-1)$
And you then get simultaneous equations
$L_x = 1+ 2\lambda x$,
$L_y = 2+ 2\lambda y$,
$L_z = 3+ 2\lambda z$,
$L_\lambda = x^2 +y^2 + z^2-1$. 
However, when I solved these simultaneously, I got some very strange solutions, so I don't know if I've done something wrong up to this point on my solving was wrong.
Thanks 
 [/INST] $x+2y+3z\leq\sqrt {x^2+y^2+z^2}\sqrt {1+4+9} = \sqrt {14}$, by Cauchy-Schwarz, so $\max$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = \dfrac{1}{\sqrt{14}}$ and $\min$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = -\dfrac{1}{\sqrt{14}}$. 
You should get a same answer if you solve the Lagrange equations correctly. 
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