I can't find the minimum point using Lagrange multipliers in that function

Question

I need to find the maximum and minimum points of the function $f(x,y,z) = x^2 + y^2 + z^2$ restricted to $x^4 + y^4 + z^4 = 1$.

I manage to find the maximum point, $\sqrt{3}$. However, the template is also pointing to a minimum point equal to 1.

I know that I can make $x = 0$ and $y = 0$ in the constraint, and thus conclude that $z= \pm 1$, or just change the combinations. However, solving the system

\begin{align*} 2x &= \lambda 4x^3 \\ 2y &= \lambda 4y^3 \\ 2z &= \lambda 4z^3 \\ x^4 + y^4 + z^4 &= 1 \end{align*}

I can't find $z= \pm 1$. Could anyone help me?

Answer 1

Your set of equations is correct. However, when solving an equation of the form $4\lambda t^3 = 2t$, you have to isolate the trivial solution $t = 0$ first and then, assuming that $t$ is non-zero (otherwise you are not allowed to divide by $t$), you can find the other solutions, namely $t = \frac{\pm1}{\sqrt{2\lambda}}$. In consequence, one has $x,y,z \in \{0,\frac{\pm1}{\sqrt{2\lambda}}\}$, with $\lambda$ being undetermined for the moment. You have thus nine different triplets $(x,y,z)$; next, you need to check whether each of them fulfills the constraint equation $-$ which will permit to find the value of $\lambda$ in most cases $-$ and finally you will have to compute the corresponding values of $f(x,y,z)$ in order to know which ones represent minima, maxima and saddle points respectively.

Answer 2

Well, Lagrange multipliers is a complicated tool while max and min can be found without calculation by writing $X=x^2$ etc. Consider the plane $a=X+Y+Z$ while $X,Y,Z\geq 0$ and belong to the unit sphere in $R^3$. The max of $a$ is for $X=Y=Z=1/\sqrt{3}$, the min for $X=1, Y=0,Z=0$ and two other points.

Of course I am not answering to your question, just complaining against academic exercises which can be solved trivially without the special techniques that the professor wants to teach.

Answer 3

From

\begin{align*} 2x &= \lambda 4x^3 \\ 2y &= \lambda 4y^3 \\ 2z &= \lambda 4z^3 \\ x^4 + y^4 + z^4 &= 1 \end{align*}

we have

\begin{align*} x^2 &= 2\lambda x^4 \\ y^2 &= 2\lambda y^4 \\ z^2 &= 2\lambda z^4 \\ x^4 + y^4 + z^4 &= 1 \end{align*}

so after addition,

$$ x^2+y^2+z^2=2\lambda $$

and after substitution

\begin{align*} x &= (x^2+y^2+z^2)x^3 \\ y &= (x^2+y^2+z^2)y^3 \\ z &= (x^2+y^2+z^2)z^3 \\ x^4 + y^4 + z^4 &= 1 \end{align*}

and we can verify that $x=y=0$ gives

\begin{align*} 0 &= 0 \\ 0 &= 0 \\ z &= z^5 \\ z^4 &= 1 \end{align*}

now as $z\ne 0$ due to the restriction, we have $z^4=1$ etc.

NOTE

Considering instead after simplifications, the system

\begin{align*} 1 &= 2\lambda x^2 \\ 1 &= 2\lambda y^2 \\ 1 &= 2\lambda z^2 \\ x^4 + y^4 + z^4 &= 1 \end{align*}

we eliminate from the set of solutions $(x=y=0,z\pm 1)$