Case Study 2 — Recovering Total Cost from Marginal Cost: A Manufacturer's Pricing Decision

Field: Economics and operations management Calculus used: Antidifferentiation to recover a total from a rate; the fixed cost as the constant of integration; initial value problems (Sections 12.8, 12.11)

The Manufacturer's Problem

A small firm makes precision aluminum brackets. Its operations manager knows something local and immediate about cost: how much it costs to make one more bracket at the current production level. Her cost accountant has measured this, run after run, and fitted a curve to it. What she has is the marginal cost — the derivative of total cost with respect to quantity. What she needs, to set a price and decide how many to produce, is the total cost itself. Getting from the one to the other is a single act of antidifferentiation, and it is the economic twin of recovering position from velocity (Section 12.11).

The logic is worth stating slowly, because the words "marginal" and "total" hide a derivative–antiderivative relationship that students often memorize without seeing. If $C(q)$ is the total cost of producing $q$ units, then the marginal cost is by definition the rate at which total cost rises as production increases:

$$C'(q) = \text{marginal cost} = \text{cost of the next unit.}$$

Marginal anything in economics is a derivative; total anything is its antiderivative. So if you know the marginal cost as a function of $q$, the total cost is

$$C(q) = \int C'(q)\,dq,$$

and — exactly as in every initial value problem — the indefinite integral hands back a whole family of cost curves, all differing by a constant. One extra fact pins down the right one. In economics that fact has a name everyone already knows: the fixed cost.

The Data and the Antiderivative

The accountant's fitted marginal-cost curve is

$$C'(q) = 100 - 0.08\,q + 0.0006\,q^2 \quad (\text{dollars per bracket}),$$

valid up to the plant's capacity of about $q = 600$ brackets per run. This shape is realistic: marginal cost starts at \$100, dips as the line warms up and workers find efficiencies (the $-0.08q$ term), then climbs again as the plant is pushed toward capacity and overtime and machine strain set in (the $+0.0006q^2$ term). The U-shaped marginal cost is one of the most common patterns in production economics.

Antidifferentiate term by term, using linearity and the power rule (Sections 12.5, 12.7):

$$C(q) = \int \left(100 - 0.08\,q + 0.0006\,q^2\right)\,dq = 100\,q - 0.04\,q^2 + 0.0002\,q^3 + K.$$

Here $K$ is the constant of integration, and so far it is undetermined — every curve $100q - 0.04q^2 + 0.0002q^3 + K$ has the same marginal cost. We have the family; we need the member.

The Fixed Cost Is the Constant of Integration

The manager knows one value of the total cost exactly: what it costs to produce nothing. Even at $q = 0$ — line idle, no aluminum bought — the firm still pays rent, insurance, and salaried staff. Suppose that fixed cost is

$$C(0) = 12{,}000\ \text{dollars per run}.$$

This is precisely an initial condition $C(0) = 12{,}000$ for the antiderivative (Section 12.8). Substitute $q = 0$ into the family:

$$C(0) = 100(0) - 0.04(0)^2 + 0.0002(0)^3 + K = K = 12{,}000.$$

So $K$ — the constant the antiderivative left free — is the fixed cost. The complete total-cost function is

$$\boxed{C(q) = 0.0002\,q^3 - 0.04\,q^2 + 100\,q + 12{,}000.}$$

That the fixed cost is the constant of integration is not a coincidence; it is forced. The constant is the value of the antiderivative when the variable is zero, and the cost when quantity is zero is exactly what "fixed cost" means. The two ideas are the same idea wearing different vocabulary.

Putting the Cost Function to Work

Now the manager can answer questions that the marginal cost alone could never answer. Take a run of $q = 500$ brackets:

$$C(500) = 0.0002(500)^3 - 0.04(500)^2 + 100(500) + 12{,}000.$$

Compute the pieces: $0.0002 \cdot 125{,}000{,}000 = 25{,}000$; $0.04 \cdot 250{,}000 = 10{,}000$; $100 \cdot 500 = 50{,}000$. So

$$C(500) = 25{,}000 - 10{,}000 + 50{,}000 + 12{,}000 = 77{,}000\ \text{dollars.}$$

The firm spends \$77,000 to make 500 brackets. From this the manager reads off the **average cost** per bracket, $C(500)/500 = \$154$, which is the floor under any sustainable price. If the market will only bear \$140 per bracket, this run loses money; if it bears \$180, the run clears \$26,000 over total cost.

Notice what the antiderivative bought her. The marginal cost at $q = 500$ is $C'(500) = 100 - 0.08(500) + 0.0006(500)^2 = 100 - 40 + 150 = \$210$ — the cost of the next bracket. That single number, useful as it is for deciding whether to make bracket number 501, says nothing about whether the run as a whole is profitable. Only the total, reconstructed by antidifferentiation, can do that. The derivative tells you about the margin; the antiderivative tells you about the whole.

A Subtlety: the Constant Cancels in Differences

Suppose instead the manager only wants the incremental cost of expanding a run from 400 to 500 brackets. She could compute $C(500) - C(400)$. Watch what happens to the fixed cost $K = 12{,}000$:

$$C(500) - C(400) = \big[\,\cdots + 12{,}000\,\big] - \big[\,\cdots + 12{,}000\,\big].$$

The $+12{,}000$ appears in both and cancels in the subtraction. The incremental cost does not depend on the fixed cost at all — which makes perfect economic sense, since rent does not change when you make a hundred more brackets. This cancellation is a small preview of something profound: in Chapter 14 the Fundamental Theorem of Calculus computes accumulated change as $F(b) - F(a)$, a difference of antiderivative values, and the "$+C$" you fought with all chapter cancels in exactly this way every time (Section 12.15). The fixed cost cancelling here is that same cancellation in miniature.

Computing the difference directly: $C(500) = 77{,}000$, and $$C(400) = 0.0002(64{,}000{,}000) - 0.04(160{,}000) + 100(400) + 12{,}000 = 12{,}800 - 6{,}400 + 40{,}000 + 12{,}000 = 58{,}400.$$ So the incremental cost of those 100 extra brackets is $77{,}000 - 58{,}400 = \$18{,}600$, or \$186 per bracket on average across the expansion — higher than the \$154 average over the whole run, because those last hundred brackets are made in the rising part of the U where marginal cost has climbed past \$200.

The Same Move, Everywhere

Strip the dollars away and the structure is identical to the braking car of Case Study 1. There we had acceleration (a rate) and recovered position (a total) by antidifferentiating, fixing the constant with an initial position. Here we have marginal cost (a rate) and recover total cost (a total) by antidifferentiating, fixing the constant with the fixed cost. The recipe — antidifferentiate, then use one known value to pin the constant — is the chapter's whole content, and it reappears whenever a quantity is known only through its rate of change (Section 12.11): charge from current, distance from speed, accumulated rainfall from rainfall rate, total revenue from marginal revenue. Economics simply runs the machine on money.

One honest caution closes the loop. This works because marginal cost is a known function of $q$ alone. If instead the rate depended on the total itself — as unconstrained growth does, where the rate of change is proportional to the current amount — plain antidifferentiation would no longer suffice, and we would need the differential-equations machinery of Chapter 19 (Section 12.11, 12.12). The manager is lucky: her cost rate depends only on how many brackets, not on how much money she has already spent, so a clean double-free antiderivative does the whole job.

Discussion Questions

  1. The constant of integration turned out to equal the fixed cost $C(0)$. Explain in your own words why this is forced rather than coincidental, referring to what the constant of an antiderivative represents.
  2. The marginal cost at $q = 500$ was \$210, but the average cost per bracket over the run was \$154. Why are these different, and which one should the manager compare against the selling price to decide whether to produce at all?
  3. When computing the incremental cost $C(500) - C(400)$, the fixed cost cancelled. Connect this to the claim in Section 12.15 that the "$+C$" never mattered for the area computation $F(b) - F(a)$.
  4. Suppose the accountant revises the fixed cost from \$12,000 to \$15,000. Which quantities in this case study change, and which (the marginal cost, the incremental cost of expansion) stay exactly the same? Why?
  5. The marginal-cost curve is U-shaped. Translate the three regions — falling, minimum, rising — into statements about the total cost curve's concavity. (Hint: marginal cost is the slope of total cost.)

Your Turn

  1. A firm has marginal cost $C'(q) = 50 + 2q$ dollars per unit and fixed cost $C(0) = 500$. Find $C(q)$, then the total cost of producing 100 units. Verify by differentiating your $C(q)$ back to $C'(q)$.
  2. With the bracket cost function above, compute $C(300)$ and the average cost per bracket at $q = 300$. Is a run of 300 cheaper per unit than a run of 500? Explain using the U-shaped marginal cost.
  3. A different firm reports marginal revenue $R'(q) = 80 - 0.5q$ and knows that revenue from selling nothing is $R(0) = 0$. Recover the total revenue function $R(q)$, and explain why, unlike cost, its constant of integration is forced to be zero.

A Short Annotated Reading List

  • Varian, H. R. (2014). Intermediate Microeconomics: A Modern Approach. Norton. The standard undergraduate treatment of cost curves; its chapters on technology and cost make the marginal-versus-total (derivative-versus-antiderivative) relationship explicit and graphical.
  • Pindyck, R. S., & Rubinfeld, D. L. (2018). Microeconomics. Pearson. Strong, applied coverage of U-shaped cost curves and the production decisions they drive, with worked numerical examples close in spirit to this case study.
  • Sydsæter, K., Hammond, P., et al. (2016). Essential Mathematics for Economic Analysis. Pearson. A bridge text that treats integration as the economist uses it — recovering totals from marginals, surplus as area — and is the natural next step once you reach the definite integral in Chapter 13.

Connections

  • Section 12.8 supplies the initial value problem (fixed cost as initial condition) at the heart of this case study.
  • Section 12.11 frames marginal-to-total cost as one instance of the universal "recover the total from the rate" pattern.
  • Section 12.15 previews the cancellation of the constant in $F(b) - F(a)$, foreshadowed here by the fixed cost cancelling in incremental-cost differences.
  • Chapter 18 returns to economic applications of the definite integral — consumer and producer surplus as areas between curves.