Using Lagrange multipliers, I need to calculate all points $(x,y,z)$ such that $$x^2-y^2+z^4$$ has a maximum or a minimum subject to the constraint that $$x^2 + y^2 + z^2 = 1$$
So, $f(x,y,z) = x^2-y^2+z^4$
and $g(x,y,z) = x^2 + y^2 + z^2 – 1$
the partial derivatives are
$$\begin{align}
2x &= 2xλ\\
-2y &= 2yλ\\
4z^3 &= 2zλ\\
\end{align}$$
The trouble i'm having is I don't know how to go further with it. Like $x = 0$ or $\lambda = 1$ , $y = 0$ or $\lambda = -1$. Can $z$ only equal $0$?
If $x = y = 0$ then $z= +/- 1$ which would mean $\lambda = 2$ . I imagine $(0,0,1)$ and $(0,0,-1)$ aren't the only critical points?
Best Answer
There are three cases
case $1$:
$$x\neq0 ~~\textrm{and}~~ y=0$$ then with $$+2x=2\lambda x$$ $\lambda$ must be $+1$. We get $x^2+z^2=1$.
case $2$:
$$x=0 ~~\textrm{and}~~ y\neq0$$ then with $$-2y=2\lambda y$$ $\lambda$ must be $-1$. We get $y^2+z^2=1$.
case $3$:
$$x=0 ~~\textrm{and}~~ y=0$$ Niether $2x=2\lambda x$ nor $-2y=2\lambda y$ restrict $\lambda$ in this case. However if $x^2+y^2+z^2=z^2=1$ then if $4z^3=2z\lambda$ only $\lambda=2$ works.