Answer
$\dfrac{L^3}{3\sqrt 3}$
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 $\nabla f=\lt yz,xz,xy \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z \gt$
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 $x=y=z$
From the given question, let us consider $g(x,y,z)=x^2+y^2+z^2=L^2$ yields $x^2+x^2+x^2=L^2$
Thus, $x=y=z=\dfrac{L}{\sqrt 3}$
Therefore, the volume of a box is $V=xyz=(\dfrac{L}{\sqrt 3})^3$
$V=(\dfrac{L}{\sqrt 3})(\dfrac{L}{\sqrt 3})(\dfrac{L}{\sqrt 3})$
$V=\dfrac{L^3}{3\sqrt 3}$
