I have seen many solutions using AM-GM relationships but I am not quite satisfied with those so here is my (incomplete) attempt at it using calculus.
I defined a plane $x+y+z=10$. The maximum volume of the cuboid lying between the axis planes and this $x+y+z=10$ plane would give us our required answer.
For a particular plane where $z$ is a constant, the plane gets intersected forming a line $x+y=a$, where $a=10-z$, hence a constant. The maximum values of $xy$ is achieved at $x=\frac{a}2, y=\frac{a}2$.
Hence our equation $x+y+z=10$ is reduced to $2x+z=10$. We can say that for maximum values of $x,y,z$ subject to aforementioned constraints, the quadrilateral plane lying between $(0,0,0), (0,0,z), (x,y,z),$ and $(x,y,0)$ will be of the maximum area, which would hence given us the maximum values as $\left(\frac{5\sqrt2}2, \frac{5\sqrt2}2, 10-5\sqrt2\right)$
So the maximum volume should be $125 - \frac{125\sqrt2}2$.
My question was whether the assumption that the vertex would lie on the line $x+y=a$ is wrong?
Instead of taking planes along $z$ axis, we could have chosen $x$ or $y$ axis, but the result would be the same so it would be fine. But there seems to be no minima in my range? How to go for it using calculus?