Marginal Analysis

Section 10.3 Marginal Analysis

Subsection 10.3.1 An Economic Model

A nice use of optimization happens in economics and finance through a technique called marginal analysis. Naively, marginal analysis is frequently presented as a sales and production question: \(C(x)\) is the cost of producing \(x\) units of a product and \(B(x)\) is the revenue from selling \(x\) units of a product. I’ll take a slightly more general interpretation, where we measure cost \(C(x)\) and benefit \(B(x)\) without specifying exactly what form those costs and benefits can take.

In this context, the derivative \(C^\prime(x)\) is called the marginal cost and the derivative \(B^\prime(x)\) is called the marginal benefit.

I can ask three questions.

First, when is there a maximum net benefit? That is, when is the difference \(B(x) - C(x)\) maximized? The derivative must be zero, so \(B^\prime(x) - C^\prime(x) = 0\text{,}\) which is \(B^\prime(x) = C^\prime(x)\text{.}\) I optimize net benefit when the marginal costs are equal. After finding such a point, I will still have to investigate to see if the point is a minimum, maximum or neither.
The second question is a local variant of the first question. If production rate is currently at some level \(x_0\) units, would more production increase the net benefit? That is, is the net benefit currently an increasing function? Increasing means positive derivative, so I look for \(B^\prime(x) - C^\prime(x) > 0\text{,}\) which is \(B^\prime(x) > C^\prime(x)\text{.}\) Increasing production increases the net benefit if the marginal benefit is larger than the marginal cost. This leads to a reasonable interpretation of these derivatives: marginal cost is (roughly) the cost to produce the next unit, and marginal benefit is the benefit due to the next unit. I can increase net benefit if the next unit has greater benefit than cost.
Lastly, I can ask a different strategic question. Instead of maximizing net benefit (like profit for a company), what if the motivation is just maximum benefit while still breaking even (more like a non-profit). This is still an optimization question, but with a different approach. Now I want to break even, which mathematically is \(B(x) = C(x)\text{.}\) The derivatives no longer come into play; I just have to solve this equality of functions and find the solution \(x\) with the highest gross benefit.

This certainly isn’t an exhaustive list of all possible questions; strategically, there could be many considerations leading to many questions. Perhaps there is a fixed budget, so the cost \(C(x)\) cannot, under any circumstances, cross a fixed maximum. If our product is a service, the goal might be maximum usage instead of maximum net benefit. Perhaps I need to minimize average cost of production instead of net benefit. Whenever I use mathematics for strategic reasons in a applied situation, it is important to remember that the mathematics only answers the questions I asked. It doesn’t tell me which question I actually want to ask, nor how to compare between the various questions.

Figure 10.3.1. Example of Marginal Analysis

Example 10.3.2.

Take the cost function \(C(x) = x^3 - x^2 + x+ 1\) and the benefit function \(B(x) = 3x\text{.}\) Assume that this model measures the produce of some objects that are then sold. \(x\) can be measured in thousands of units and \(B\) and \(C\) can be measured in millions of dollars. Notice that \(C(0) = 1\text{,}\) which could represent an initial start-up cost before the production of a single unit.

I calculate the derivatives: \(C^\prime(x) = 3x^2 -2x+1\) and \(B^\prime(x) = 3\text{.}\) These are equal when \(x = \frac{1}{3} (1 \pm \sqrt{7})\text{.}\) Discarding the negative, the other root is approximately \(1.215\text{,}\) so a production of 1215 units gives the maximum net benefit. Below that point, marginal benefit exceeds cost, so I should increase production. Above that point, marginal cost will exceed benefit, so I should decrease production.

The break even points are found by solving the cubic: \(x^3 -x^2-2x+1 - 3x = 0\text{.}\) The cubic doesn’t factor nicely, but a computer can give the approximate even \(x\) values: \(-1.25\text{,}\) \(0.45\) and \(1.81\text{.}\) I discard the negative value again. The largest break-even point is at approximately \(1.81\text{.}\) All these values and the behaviour of the cost and benefit functions are shown in Figure 10.3.1.

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