Question:

By using Lagrange multipliers, find the extrema of \(x^2+2y^2+3z^2\) when \(3x+2y+z=-1\).

Answer:
This is the exactly the same question as previously provided.
But this time, using a Lagrange multiplier, we solve the problem.
Now, define the following function:

\[ F(x,y,z) = x^2+2y^2+3z^2 – \lambda (3x+2y+z+1) \] where \(\lambda\) is called the Lagrange multiplier.

Take each partial derivative.

\[ F_x = 2x-3\lambda = 0 \]

\[ F_y = 4y-2\lambda = 0 \]

\[ F_z = 6z-\lambda = 0 \]

\[ F_{\lambda} = 3x+2y+z+1 = 0 \]

There are four unknowns and four equations. Solve for each variable.

\[ x=-9/34 \]

\[ y=-3/34 \]

\[ z=1/34 \]

\[ \lambda = -6/34 \]

If you plug \(x\), \(y\), \(z\), and \(\lambda\), we obtain \(F = 3/34\) as the extremum.