Question

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If a value does not exist, enter NONE.) f(x,y,z) = x2y2z2; x2 + y2 + z2 = 1

Answer 1

The Lagrangian is

$L(x,y,z,\lambda)=x^2y^2z^2+\lambda(x^2+y^2+z^2-1)$

where we assume $\lambda\neq0$, with critical points wherever the partial derivatives are identically zero:

$L_x=2xy^2z^2+2\lambda x=0\implies2x(y^2z^2+\lambda)=0\implies x=0\text{ or }y^2z^2=-\lambda$

(note that the latter case requires $\lambda$, and we'll see the same requirement in the next two equations)

$L_y=2x^2yz^2+2\lambda y=0\implies2y(x^2z^2+\lambda)=0\implies y=0\text{ or }x^2z^2=-\lambda$

$L_z=2x^2y^2z+2\lambda z=0\implies2z(x^2y^2+\lambda)=0\implies z=0\text{ or }x^2y^2=-\lambda$

$L_\lambda=x^2+y^2+z^2-1=0$

If either $x=0$, $y=0$, or $z=0$, then we can throw out the latter cases for $L_x=0$, $L_y=0$, and $L_z=0$, since we don't want $\lambda=0$. If, for instance, we pick $x=y=0$, then we're left with $z^2=1\implies z=\pm1$, for which $f(x,y,z)=0$. This means we have 6 critical points $(x,y,z)$ where two of the coordinates are 0, and the remaining one is either 1 or -1.

In the latter case,

$y^2z^2=x^2z^2=-\lambda\implies x^2=y^2\implies x=\pm y$

$x^2z^2=x^2y^2=-\lambda\implies z^2=y^2\implies y=\pm z$

$y^2z^2=x^2y^2=-\lambda\implies x^2=z^2\implies z=\pm x$

If $x=y=z$, then

$3x^2=1\implies x=y=z=\pm\dfrac1{\sqrt3}$

A similar thing happens in all the other cases, which leads us to getting 8 possible critical points, $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$; at each of these points, we have $f(x,y,z)=\dfrac1{27}$.

To summarize: $f(x,y,z)$ has a maximum value of 1/27 and a minimum value of 0.