Consider the following pair of a priori independent problems...
A manufacturer's revenue is $\$100s^{\frac{1}{3}}h^{\frac{2}{3}}$, where $s$ is the number of tons of steel they purchase and $h$ is the number of hours of labor they hire. Steel costs $\$2000$ per ton, labor costs $\$20$ per hour, and their budget is $\$20000$. Find their maximum revenue without exceeding their budget, and the corresponding $s$, $h$, and $\lambda$.
In other words...$$\max_{s,\ h}\ [100s^{\frac{1}{3}}h^{\frac{2}{3}}]$$$$2000s + 20h \leq 20000$$$$s \geq 0$$$$h \geq 0$$
A manufacturer's revenue is $\$100s^{\frac{1}{3}}h^{\frac{2}{3}}$, where $s$ is the number of tons of steel they purchase and $h$ is the number of hours of labor they hire. Steel costs $\$2000$ per ton, labor costs $\$20$ per hour, and their revenue target is $\$\frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$. Find their minimum budget without underperforming their revenue target, and the corresponding $s$, $h$, and $\lambda$.
In other words...$$\min_{s,\ h}\ [2000s + 20h]$$$$100s^{\frac{1}{3}}h^{\frac{2}{3}} \geq \frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$$$$s \geq 0$$$$h \geq 0$$
Reasoning that the inequality in the budget constraint/revenue target constraint hold with equality and using the method of Lagrange Multipliers, the solution to the first problem is $\$\frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$ with $s = \frac{10}{3}$, $h = \frac{2000}{3}$, and $\lambda = \frac{\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$, and the solution to the second problem is $\$20000$ with $s = \frac{10}{3}$, $h = \frac{2000}{3}$, and $\lambda = \frac{3}{\sqrt[\leftroot{-2}\uproot{3}3]{5}}$.
Of course, these two problems were not actually written independently. The revenue target value $\frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$ I put into the second problem statement was inspired by the solution to the first problem. However, that the original budget value $20000$ was recovered in the solution to the second problem, and that $s$ and $h$ match in both problems, were not reverse engineered. Maximizing on the revenue surface and minimizing on the budget surface are clearly strongly coupled. On multiple occasions, while and after I studied Multivariable Calculus, I have been informed that this is an example of duality (one such instance).
$1)$ But now I am studying linear programming, and this feels strangely both concordant and discordant with the concept of LP duality. The coupling between maximization on an objective function surface and minimization on a constraint surface (or vice-versa) is the soul of duality. However, in LP duality I have always seen this play out in terms of a new set of independent variables and bounded (weak duality) or equal (strong duality) objective function values. The above pair of problems is of a slightly different nature.
One idea I had was that the expressions in the above problems may be solved in terms of the "wrong" variable, relative to the ordinary duality paradigm. For example, if revenue is $R = f(s,\ h)$, perhaps I would get a more recognizable duality structure by considering $s = g(R,\ h)$, where $g$ is the inverse of $f$ while $h$ is treated as a constant (similarly for the budget and alternatively for $h$ instead of $s$). However, at least while $h$ is treated as a constant, the existence of an extremum in $R = f(s,\ h)$ entails the non-existence of any extrema in $s = g(R,\ h)$ (unless multifunctions are amenable to duality, but we can bracket that question). This is clear from thinking about the reflection of an inverse function in $2$ dimensions. I can't quite convince myself whether this continues to hold in the broader $3$-dimensional space once $h$ is no longer treated as a constant (that is, in general, the proposition that for any $z = h(\vec x)$, at most either $z$ or a single $x_i$ solved in terms of everything else can have any extrema).
Obviously, the above problems are not linear, thus they should not literally conform to the LP duality paradigm, and I have not yet studied any nonlinear programming, so perhaps there are simply more than a single pair of dual maximization-minimization couplings which arise out of nonlinearity? How does the above pair of problems fit into the duality paradigm?
$2)$ How does the Lagrange Multiplier $\lambda$ fit into this paradigm? I have seen a variable $\lambda$ used in what appears to be applying the minimax theorem to the Lagrangian as a representation of duality, but in the above problems, $\lambda$ is just an artifact of the Lagrange Multiplier solution method which turns out to have an interpretation as a certain derivative. It is not a quantity to be optimized in either the maximization or minimization problem. Beyond the surprise derivative interpretation, it does not appear to be an intrinsic variable in either problem, and certainly does not appear symmetric to the number of tons of steel or number of hours of labor, as the primal and dual variables in the zero-sum game application of the minimax theorem symmetrically represent each player's mixed strategy probability vector. On the other hand, $\lambda$ is not such a common variable name, and putting it into a Lagrangian would be a misleading choice if it was not somehow connected to the Lagrange Multiplier $\lambda$. What is going on here?
EDIT: I may have discovered another instance with this (or similar) structure. The fractional set cover problem and a similarly fractional version of the maximum coverage problem seem analogous to my second and first problems, respectively. "Maximize revenue subject to the budget by varying $s$ and $h$" becomes "maximize area covered subject to the subset budget by varying the subset activations (possibly fractionally)." "Minimize cost subject to the revenue target by varying $s$ and $h$" becomes "minimize subset activation subject to the area coverage target by varying the subset activations (possibly fractionally)." An answer in terms of the relationships between coverage, set cover, and each of their dual linear programs may also be helpful.