#### Answer

$(\pm 3^{-1/4},3^{-1/4}\sqrt 2,\pm 3^{1/4}),(\pm 3^{-1/4},-3^{-1/4}\sqrt 2,\pm 3^{1/4})$

#### Work Step by Step

Need to apply Lagrange Multipliers Method to determine the dimensions of a rectangular box of maximum volume.
we have $\nabla f=\lambda \nabla g$
The volume of a box is $V=xyz$
or, $f=V=xyz$
Consider $\nabla f=\lt 2x,2y,2z \gt$ and $\lambda \nabla g=\lambda \lt y^2z^3,2xyz^3,3xy^2z^2 \gt$
Using the constraint condition we get, $yz=\lambda 2x, xz=\lambda 2y,xy=\lambda 2z$
Simplify to get the value of $x,y$ and $z$
We have $2x^2=y^2,3y^2=2z^2,xz=1$
From the given question, let us consider $g(x,y,z)=xy^2z^3$ yields $x=\pm \dfrac{1}{3^{1/4}}$
Thus, $y=\pm \dfrac{\sqrt 2}{3^{1/4}}$
and $z=\pm 3^{1/4}$
Hence, the points which are closet to the origin are:
$(\pm 3^{-1/4},3^{-1/4}\sqrt 2,\pm 3^{1/4}),(\pm 3^{-1/4},-3^{-1/4}\sqrt 2,\pm 3^{1/4})$