This is just something I'm trying to do as a "build up" to a three dimensional equivalent, to see how it might work.
Consider a quadrilateral with vertices $(0,0), (x, 0), (0, y)$ and $(x,y)$ where this last vertex lies on the line $\frac{x}{a} + \frac{y}{b} = 1$. I am trying to maximise the area of this quadrilateral which I presume will be given in terms of $a$ and $b$. Let $A(x, y) = xy$ determine the area of the quadrilateral and let our constraint be given by $C(x, y) = \frac{x}{a} + \frac{y}{b} - 1$. Then we can form a new function $F(x, y, \lambda) = xy + \lambda\left( \frac{x}{a}+ \frac{y}{b} - 1\right).$ Now, to maximise we want
\begin{align} F_x &= y + \frac{\lambda}{a} = 0;\\ F_y &= x + \frac{\lambda}{b} = 0;\\ F_\lambda &= \frac{x}{a} + \frac{y}{b} - 1 = 0. \end{align}
From $F_x$ and $F_y$ we have that $y = -\frac{\lambda}{a}$ and $x = -\frac{\lambda}{b}$ respectively. Passing these into $F_\lambda$ we obtain
\begin{align} F_\lambda &= -\frac{\lambda}{ab} -\frac{\lambda}{ab} - 1\\ &= -\frac{2\lambda}{ab} - 1\\ &= 0 \end{align}
and so $\lambda =-\frac{ab}{2}$
Now that we have $\lambda$ we can determine the values $y = \frac{b}{2}$ and $x = \frac{a}{2}$ and these are the values of $x$ and $y$ that maximise the area of this quadrilateral under $\frac{x}{a} + \frac{y}{b} = 1$.
Does this make sense and can I extrapolate this kind of process to a three dimensional problem involving a regular cuboid underneath a plane?
Your approach might work, but will need some modifications.
In general, to apply Lagrange, you need to ensure that a solution exists, but this fails when you go to large dimensions.
Consider $n=3$, with the problem $\min \{x_1 x_2 x_3 | \sum_k x_k = 1 \}$.
If we choose the point $(-1,y,2-y)$, it is easy to verify that this is satisfies the constraints, and has cost $y^2-2y$, hence is unbounded above and hence no $\max$ exists.
If you restrict the domain by requiring $x_k \ge 0$, then the Lagrange approach will work, but you need to show that the non negativity constraints are not active at a solution.