I have been experimenting with the following problem paraphrased from Khan Academy:

A manufacturer's revenue is $100h^{2/3}s^{1/3}$, where $h$ is the number of hours of labor hired, and $s$ is the number of tons of steel purchased, and their budget is $\$20000$. Labor costs $\$20$ per hour and steel costs $\$2000$ per ton. Maximize their revenue.

So the problem can be summarized as $$\begin{cases}R(h,\ s) = 100h^{2/3}s^{1/3} =\ ????\ \Tiny(objective\ function) \\ B(h,\ s) = 20h + 2000s = 20000\ \Tiny(constraint)\end{cases},$$ and perhaps we're interested in the maximal value of the objective function along with the associated values of $h$, $s$, and $\lambda$. The answer is $h = \frac{2000}{3}$, $s = \frac{10}{3}$, $\lambda = \frac{\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$, $R = \frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$.

Let's now forget the values for $h$, $s$, $\lambda$, and the original constraint, but remember that $R = \frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$. We can solve the following problem:

A manufacturer's revenue target is $\$\frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$, given by $100h^{2/3}s^{1/3}$, where $h$ is the number of hours of labor hired, and $s$ is the number of tons of steel purchased. Labor costs $\$20$ per hour and steel costs $\$2000$ per ton. Minimize the budget while achieving the target.

Now the problem is $$\begin{cases}R(h,\ s) = 100h^{2/3}s^{1/3} =\ \frac{20000\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}\ \Tiny(constraint) \\ B(h,\ s) = 20h + 2000s =\ ????\ \Tiny(objective\ function)\end{cases},$$ and the answer is $h = \frac{2000}{3}$, $s = \frac{10}{3}$, $\lambda = \frac{\sqrt[\leftroot{-2}\uproot{3}3]{5}}{3}$, $B = 20000$.

So we've recovered the original constraint, and rediscovered $h$, $s$, and $\lambda$. This provides an insightful, higher-level perspective on the problem. We can now see that it is somewhat arbitrary which function is the constraint, and which is the objective function; the choice depends only on what is known about the real-world problem.

Drawing upon this higher-level perspective, my next idea was to maximize the manufacturer's profit. The first problem was about finding the best contour line of $R$ given a particular contour line of $B$, while the second was about finding the best contour line of $B$ given a particular contour line of $R$. Why not let both vary and find the overall best combination of the two contour lines?

My approach was to define $P(h,\ s)$ as $100h^{2/3}s^{1/3} - 20h - 2000s$, maintaining the assumption that whatever budget we derive, we're going to spend all of it, so that budget and cost remain synonymous. This is an unconstrained optimization problem (NOTE: introducing the Lagrangian is one way to turn a constrained optimization problem into an unconstrained optimization problem, but I think this approach is different, as we're now looking at a higher-level problem, whereas the Lagrangian would only serve as new notation to express one of the two previous problems). Beginning the unconstrained optimization solution process, we see that $\nabla P(h,\ s) = \vec{0} \implies \begin{bmatrix}\frac{200s^{1/3}}{3h^{1/3}} - 20 \\ \frac{100h^{2/3}}{3s^{2/3}} - 2000\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix} \implies \begin{bmatrix}\frac{200s^{1/3}}{3h^{1/3}} \\ \frac{100h^{2/3}}{3s^{2/3}}\end{bmatrix} = \begin{bmatrix}20 \\ 2000\end{bmatrix}$. Pausing here, it is very interesting to see that this is equivalent to both $\nabla R(h,\ s) = \lambda\nabla B(h,\ s)$ and $\nabla B(h,\ s) = \lambda\nabla R(h,\ s)$, where $\lambda = 1$. We still have, in a certain sense, a constrained optimization problem. Furthermore, we've gained one degree of freedom by essentially having two objective functions and no constraint, but lost one degree of freedom by pinpointing $\lambda$. I'm guessing this discovery of $\lambda$ is related to the newfound symmetry between $R$ and $B$; since we aren't equipping one or the other with a constraint value, $\lambda$ must equal its own multiplicative inverse to satisfy both $\nabla R(h,\ s) = \lambda\nabla B(h,\ s)$ and $\nabla B(h,\ s) = \lambda\nabla R(h,\ s)$.

Since our net degrees of freedom remain unchanged, we still have a chance of finding one or a few solutions. Alas, however, this system has no non-trivial solutions. It would appear that Khan's widget manufacturer's optimal strategy is to hire $0$ hours of labor and purchase $0$ tons of steel.

I have two related questions to ask, aiming for a deeper understanding of what's going on here. Firstly, if the expressions for $R$ and $B$ were such that $P$ did have a non-trivial maximum, would it, and/or the associated $h$ and $s$ values, be related to those of the initial pair of problems? I would not expect to find anything interesting here, since the original budget value of $20000$ was arbitrary, but generating and solving these problems presently takes me far too long to test this hypothesis.

Secondly, when constructing the third, higher-level problem, we utilized the real-world observation that profit = revenue - cost, a relationship which would not hold between objective functions and constraints in general. However, we obtained a very mathematically clean result, which unexpectedly (at least to me) invited us to decompose profit, and reconsider revenue and cost as separate entities once again. Furthermore, we stumbled upon the value of $\lambda$, which made the degrees of freedom work out. Finally, this value of $\lambda$ was the multiplicative identity, which allowed us to preserve the symmetricity in our view of the relationship between $R$ and $B$, with respect to which is the objective function and which is the constraint. Are these extremely happy coincidences, or is the profit, revenue, and cost model an example of a purer mathematical principle, one deeply intertwined with optimization problems?

  • $\begingroup$ If I understand your question, one way to look at it is that profit maximization can be decomposed into two parts. For a given revenue, determine the optimal mix of inputs (or for a given total cost of inputs, maximize revenue). Then find the revenue that maximized profits, given that you have already found the cheapest way to produce that level of revenue. $\endgroup$ – Trurl Dec 21 '18 at 14:10

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