Finding the maximum volume of a tetrahedron with 3 concurrent edges Can someone help me out? I am not good at math. Thank you.

Find the maximum volume of the tetrahedron that have three concurrent edges and satisfy the following condition:
The sum of three edges is constant.
One edge is double the length of another edge.

I found a formula to calculate an irregular tetrahedron without having to know the height but I have no idea how to know which volume is larger since the given formula is in a matrix format. Am I heading to the right direction on this question?
The source of formula (very new here, don't know how to insert matrix):
http://mathforum.org/dr.math/faq/formulas/faq.irreg.tetrahedron.html#:~:text=It%20is%20irregular%20if%20and,not%20all%20of%20equal%20measure.
 A: If you operate with the relations between $x_1,x_2,x_3$ you are given, you can define the volume $V$ in function of $x_2$ this way:
$$
x_1=2x_2 \phantom{a},\phantom{a} x_2=x_2 \phantom{a},\phantom{a} x_3=L-3x_2.
$$
So you en up with the volume being (it is $V=\frac{1}{6}x_1x_2x_3$ because that's the maximal volume formula for the irregular tetrahedron, with the edges making right angles between them):
$$
V(x_2)=\frac{1}{6}x_1\cdot x_2\cdot x_3=\frac{1}{3}Lx^2_2-x^3_2.
$$
If you derivate this expression, you end up with
$$
V'(x_2)=x_2\left(\frac{2L}{3}-3x_2\right).
$$
The equation $V'(x_2)=0$ gives you two solutions, one is obviously $x_2=0$, but that's the minimun, not what we want. The other solution is the maximun, which is $\boxed{x_2=\frac{2L}{9}}$.
This gives us the final results:
$$\boxed{x_1=\frac{4L}{9},\quad x_2=\frac{2L}{9},\quad x_3=\frac{L}{3}}.$$
$$\boxed{V=\frac{1}{6}\frac{8L^3}{3^5}=\frac{4L^3}{3^6}}$$
Tell me if there's anything you don't understand from my solution, I'll try to explain it to you.
A: Let us consider a reference system in which the concurrent edges are located in the origin $O$ of the reference, the $OA=x_1$ edge lays on the $x$-axis, the $OB=x_2$ edge lays on the $xy$ coordinate plane and the $OC=x_3$ edge is located in the half-space $z>0$, as in the figure.
Let's call $\alpha$ the angle between the edges $x_1$ and $x_2$ and $\beta$ the angle between the edge $x_3$ and the plane $xy$.

Then
\begin{align}
V(x_2,\alpha,\beta) &= \frac{1}{3}Ah=\frac{1}{3}\left[\frac{1}{2}\overline{OA}\cdot\overline{OB}\sin\alpha\right]OC\sin\beta\\
  &= \frac{1}{3}\left[\frac{1}{2}2x_2\cdot x_2\sin\alpha\right](L-3x_2)\sin\beta\\
  &= \frac{1}{3}x_2^2(L-3x_2)\sin\alpha\sin\beta
\end{align}
then, to look for the maximum we have
\begin{align}
  \frac{\partial V}{\partial x_2}   &= \frac{1}{3}x_2(2L-9x_2)\sin\alpha\sin\beta\\
  \frac{\partial V}{\partial\alpha} &= \frac{1}{3}x_2^2(L-3x_2)\cos\alpha\sin\beta\\
  \frac{\partial V}{\partial\beta}  &= \frac{1}{3}x_2^2(L-3x_2)\sin\alpha\cos\beta
\end{align}
from which we have $\alpha=\beta=\pi/2$ and $x_2=2L/9$.
