Find the extrema: an optimization problem with Lagrange multipliers

Question:
By using Lagrange multipliers, find the extrema of $x^2+2y^2+3z^2$ when $3x+2y+z=-1$.

Answer:
Using a Lagrange multiplier, we define the following function:
\[ F(x,y,z) =  x^2+2y^2+3z^2 - \lambda (3x+2y+z+1) \]
Take each partial derivative.
\[ F_x = 2x-3\lambda = 0  \\
F_y = 4y-2\lambda = 0    \\
F_z = 6z-\lambda = 0   \\
F_{\lambda} = 3x+2y+z+1 = 0   \]
They have four unknowns and four equations. Solve for each variable.
\[ x=-9/34  \\
  y=-3/34  \\
  z=1/34  \\
  \lambda = -6/34 \]
If you plug $x$, $y$, and $z$, we obtain $F = 3/34$ as the extremum.

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