# CoV2

Find the extrema of $x^{2}+y^{2}+z^{2}\,$ subject to the constraint $x^{2}+2y^{2}-z^{2}-1=0\,$.

Using Lagrange multipliers, make a new function:

$f=x^{2}+y^{2}+z^{2}+\lambda x^{2}+2\lambda y^{2}-\lambda z^{2}-\lambda =0\,$

Take the partial derivatives of this function with respect to $x$, $y$, and $z$.

1) ${\frac {\partial f}{\partial x}}=2x+2\lambda x=0\,$

2) ${\frac {\partial f}{\partial y}}=2y+4\lambda y=0\,$

3) ${\frac {\partial f}{\partial z}}=2z-2\lambda z=0\,$

For each equation above, find a value of lambda that works and force that value into the other two equations. Choose the other two variables appropriately to satisfy the equalities.

From 1), $\lambda =-1$, and in that case $y=z=0$. Plugging into our constraint equation yields $x=1,-1$.

From 2), $\lambda =-1/2$, and in that case $x=z=0,y=1/{\sqrt {2}},-1/{\sqrt {2}}$.

From 3), $\lambda =1$, and in that case $x=y=0,z=i,-i$.

Combining these results, the extrema are

$(1,0,0),(-1,0,0),(0,1/{\sqrt {2}},0),(0,-1/{\sqrt {2}},0),(0,0,i),(0,0,-i)\,$

The last two solutions may be disregarded if we are restricting to real variables.