such a box has one vertex at the origin (0, 0, 0), and the vertex opposite this one is affixed to the plane with coordinates . the volume of the box is then , which we want to maximize subject to the constraint . of course, we don't want a degenerate box, so we assume each of is positive.
we can use lagrange multipliers - the lagrangian is
with partial derivatives (set equal to 0)
so the largest volume that can be attained is .
let us use engineering notation, for the sake of shortening the zeros expression, so 100,000 is just 1e5, namely 1 followed by 5 zeros, and 399 million is just 399e6, or 399 followed by 6 zeros and so on.
-30 sqrt50 x^5.
= 2 sqrt 5x^3* -3 sqrt10 x^2
= -6 sqrt50 x^5