Question

use lagrange multipliers to find the absolute extrema of f(x,y,z)= xyz subject to the constraint x^2+2y^2+4z^2=4.

Answer 1

\[yz= 2x \lambda \]

\[xz= 4y \lambda\]

\[xy= 8z \lambda\]

i always end up with incomplete critical points when solving problems like this..

Answer 2

\[\begin{cases}yz=2x\lambda\\
xz=4y\lambda\\
xy=8z\lambda\\
x^2+2y^2+4z^2=4\end{cases}\]
Multiply the first equation by \(x\), the second by \(y\), and the third by \(z\):
\[\begin{cases}xyz=2x^2\lambda\\
xyz=4y^2\lambda\\
xyz=8z^2\lambda\\
x^2+2y^2+4z^2=4\end{cases}\]
So now you have
\[2x^2\lambda=4y^2\lambda=8z^2\lambda\\
x^2=2y^2=4z^2\]
(assuming \(\lambda\not=0\)).
Substituting into the fourth equation:
\[x^2+x^2+x^2=4\\
x^2=\frac{4}{3}\\
x=\pm\sqrt\frac{4}{3}\]
So you also have
\[2y^2=\frac{4}{3}\\
y=\pm\sqrt\frac{2}{3}\]
and
\[4z^2=\frac{4}{3}\\
z=\pm\sqrt\frac{1}{3}\]