Minimize $x^2-y^2+z^4$ subject to $x^2 + y^2 + z^2 = 1$

Question

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?

Answer 1

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.

Answer 2

You need a systematic way to separate the cases:

Take the equation $2x=2x\lambda$ first. OK, obviously it's the same as $x=x\lambda$, which is the same as $x(1-\lambda)=0$, and this means either $x=0$ or $\lambda=1$.

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So, case 1: $\lambda=1$

In this case, $-2y=2y\lambda = 2y$ meaning $y=-y$ and $y=0$.

Also, $4z^3=2z\lambda=2z$ meaning $2z^3=z$ or $z(2z^2-1)=z(\sqrt{2}z-1)(\sqrt{2}z + 1) = 0$. This gives three possibilities for $z$, because either of the three factors could be $0$.

Once you have $y$ and $z$, you can still calculate $x$ from $x^2+y^2+z^2=1$.

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Case $2$: $x=0$.

In this case, the equation $2x=2x\lambda$ is $0=0$ and we can't get anything from it. So, you can focus on the second equation:

$-2y = 2y\lambda$, which simplifies to $y(\lambda+1)=0$.

Again, you can separate two sub-cases:

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Case $2.1$:

$y=0$. If $y=0$, and you already know that $x=0$ (because we are still in the case $x=0$, we just took another sub-case), you know that $z=\pm 1$ from $x^2+y^2+z^2=1$.

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Case $2.2$:

$\lambda = -1$:

In this case, $2x=2x\lambda=-2x$ which means that $x=-x$, so $x=0$.

The third equation becomes $4z^3=-2z$ which is the same as $z(2z^2+1)=0$. Because $2z^2+1>0$, this equation has only one solution.

Again, you can calculate $y$ in this cas from $x^2+y^2+z^2=1$.