$$\begin{array}{ll} \text{extremize} & xy+2yz+3zx\\ \text{subject to} & x^2+y^2+z^2=1\end{array}$$
How to find the maximum/minimum using Lagrange multipliers?
Context: This is not a homework problem, my friend and I often make up problems to challenge each other. We both love Maths and we are both students.
I have improved my answer based on user247327's suggestion, and I have found the maximum value of 2.056545.Thank you for contributing ideas to my questions.
On
$$ x y + 2 y z + 3 z x = \begin{bmatrix} x \\ y \\ z\end{bmatrix}^\top \begin{bmatrix} 0 & \frac12 & \frac32 \\ \frac12 & 0 & 1 \\ \frac32 & 1 & 0\end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} \leq \lambda_{\max } \begin{bmatrix} 0 & \frac12 & \frac32 \\ \frac12 & 0 & 1 \\ \frac32 & 1 & 0\end{bmatrix} \approx \color{blue}{2.05655}$$
Using "Lagrange multipliers" you look at $F(x)= xy+ 2yz+ 3xz+ \lambda(x^2+ y^2+ z^2- 1)$.
Then $F_x= y+ 3z+ 2x\lambda= 0$ $F_y= x+ 2z+ 2y\lambda= 0$ $F_z= 2y+ 3x+ 2z\lambda= 0$ and $F_\lambda= x^2+ y^2+ z^2- 1= 0$.
Since a value for $\lambda$ is not required for the answer, I would eliminate it from the first three equation. Multiply both sides of the first equation by y to get $y^2+ 3yz+ xy\lambda= 0$ and multiply both sides of the second equation by x to get $x^2+ 2xz+ xy\lambda= 0$. Now subtract: $y^2- x^2+ 3yz- 2xz= 0$.
Similarly, multiply the first equation by z to get $yz+ 3z^2+ 2xz\lambda= 0$ and multiply the third equation by x to get $2xy+ x^2+ 2xz\lambda= 0$. Now subtract: $yz- 2xy+ 3z^2- x^2= 0$.
We now have three equations $y^2- x^2+ 3yz- 2xz= 0$ $yz- 2xy+ 3z^2- x^2= 0$ and $x^2+ y^2+ z^2= 1$ to solve for x, y, and z.