From Marginal Cost to Total Cost
A production model may describe how total cost changes as output changes, without giving the total cost directly. That rate of change is called marginal cost. To recover total cost from it, we also need a baseline: the cost at a specified production level.
This is the same rate-and-accumulation structure used in other models: a rate describes change, while an initial or reference value determines the level of the accumulated quantity. Here the changing variable is production, and the accumulated quantity is cost.
In a continuous model, marginal cost describes the instantaneous change in total cost as production changes. It is not itself the total cost, nor is it generally the cost of producing every item up to \(q\). A baseline is needed to recover the total-cost level.
The Accumulation Formula
Suppose the marginal-cost function is \(m(q)\), so that \(C'(q)=m(q)\), and suppose the cost at a reference production level \(q_0\) is known: \(C(q_0)=C_0\). If \(m\) is continuous between \(q_0\) and \(q\), the Fundamental Theorem of Calculus gives the cost function:
The variable \(x\) is a dummy variable of integration: it represents production levels along the interval, while \(q\) is the endpoint at which we want to find the cost. The definite integral gives the change in cost from \(q_0\) to \(q\). Adding \(C_0\) gives the total cost at \(q\).
If \(q<q_0\), the integral is oriented in the usual way: reversing the endpoints changes its sign. This makes the formula consistent with cost differences, since \(C(q)-C(q_0)\) is also the negative of \(C(q_0)-C(q)\).
Why the Formula Works
The formula follows from the relationship between derivatives and definite integrals. The argument also shows why the baseline is essential.
Define \(G(q)=C_0+\int_{q_0}^{q}m(x)\,dx\). This function has the specified baseline because the integral from \(q_0\) to itself is zero, so \(G(q_0)=C_0\).
Since \(m\) is continuous, the Fundamental Theorem of Calculus gives \(G'(q)=m(q)\).
If \(C'(q)=m(q)\), then \((C-G)'(q)=0\). By the Mean Value Theorem, \(C-G\) is constant on the interval.
At \(q_0\), \(C(q_0)-G(q_0)=C_0-C_0=0\). Thus the constant is zero, \(C(q)=G(q)\), and the accumulation formula follows.
This proof requires the stated differentiability and continuity conditions on the interval. If marginal cost is piecewise continuous, the integral can still be evaluated piece by piece. The resulting cost is continuous across a finite rate-change boundary, but its derivative at that boundary exists only if the marginal-cost values from both sides agree.
Worked Example: A Smooth Marginal-Cost Model
Worked Example: A Smooth Marginal-Cost Model
A manufacturer has marginal cost \(m(q)=12+0.04q\) dollars per item, where \(q\) is the production level. Fixed and setup costs give the baseline \(C(0)=500\) dollars. Find the total-cost function and the cost accumulated as production increases from 20 to 50 items.
Use the baseline and integrate marginal cost from \(0\) to \(q\):
Differentiate to check the recovered marginal cost:
For the interval from 20 to 50 items, calculate the accumulated cost change directly:
Check using total costs at the endpoints:
The definite integral gives the increase in cost over those 30 items. The baseline \(500\) dollars is not part of that increase; it is included in both endpoint costs and cancels in their difference.
Piecewise Marginal Cost and Rate Changes
Marginal cost may change when production enters a different operating regime. In that case, integrate each formula over the portion of the production interval where it applies, then add the changes. A single value assigned to marginal cost at one boundary does not affect a definite integral. However, the derivative of total cost at that boundary depends on the rates on both sides.
Worked Example: Two Production Regimes
Worked Example: Two Production Regimes
Suppose a plant has \(C(0)=200\) dollars and the following marginal-cost model:
The marginal-cost model does not specify a derivative at \(q=40\). Find the total cost at \(q=50\), and find the cost accumulated as production increases from 30 to 50 items.
First integrate the lower-production rate from 0 to 40. Its cost increase is
Therefore the cost at the boundary is \(C(40)=200+400=600\) dollars. For production above 40, integrate the second rate from 40 to \(q\):
At 50 items, this gives
Now calculate the increase from 30 to 50 by splitting at 40. From 30 to 40, the first regime contributes
From 40 to 50, the second regime contributes
Thus the accumulated cost over the full interval is \(115+185=300\) dollars. Check it using endpoint costs: \(C(30)=200+8(30)+0.05(30)^2=200+240+45=485\) dollars, so \(C(50)-C(30)=785-485=300\) dollars, as required.
A Boundary Pitfall: Continuous Cost Does Not Guarantee a Derivative
In the piecewise example, cost is continuous at 40 items: both formulas give \(C(40)=600\). But it would be incorrect to claim that the derivative there equals both marginal-cost formulas, or that the cost function is differentiable there. The rates immediately to the left and right differ.
To check directly, let \(h<0\) be a small change from 40. The first regime gives \(C(40+h)-C(40)=12h+0.05h^2\), so the left-hand difference quotient tends to 12. For \(h>0\), the second regime gives \(C(40+h)-C(40)=18h+0.05h^2\), so the right-hand difference quotient tends to 18. Since these limits are unequal, \(C'(40)\) does not exist.
Worked Example: Recovering Cost from a Nonlinear Rate
Worked Example: Recovering Cost from a Nonlinear Rate
Suppose marginal cost is \(m(q)=3q^2-4q+10\) dollars per item and the cost at \(q=2\) is \(C(2)=150\) dollars. Find the cost function and the cost increase from 2 to 5 items.
Integrate from the known baseline level, retaining the baseline cost:
The primitive at 2 is \(2^3-2(2^2)+10(2)=8-8+20=20\). Therefore
Differentiate to verify the marginal-cost model, and substitute the baseline to check its value:
The accumulated cost from 2 to 5 items is
As an endpoint check, \(C(5)=125-50+50+130=255\) dollars and \(C(5)-C(2)=255-150=105\) dollars. The integral and the difference in total costs agree.
Units and Interpretation
Units provide a useful check on the model. If \(q\) is measured in items and \(C(q)\) in dollars, then \(C'(q)\) is measured in dollars per item. In an integral such as \(\int_{q_0}^{q}m(x)\,dx\), multiplying dollars per item by a change measured in items gives dollars, the units of a cost change.
For a production interval \([q_0,q_1]\), the integral accumulates the marginal cost across that interval. If marginal cost is nonnegative throughout, the accumulated cost is nonnegative when \(q_1\ge q_0\). More generally, the signed integral records the net change in cost, consistent with the endpoint difference \(C(q_1)-C(q_0)\).
The baseline determines which total-cost function is appropriate. Marginal cost alone specifies changes in cost but cannot determine the cost level: adding a constant to a cost function leaves its derivative unchanged. A known baseline fixes that constant. When \(q_0=0\), \(C(0)\) commonly represents fixed or setup costs in the model, so omitting it would give variable cost rather than total cost.
Check Your Understanding
Use the accumulation formula and check your answers against the meaning of the baseline.
- If \(C'(q)=5+0.2q\) and \(C(0)=100\), write the total-cost function.
- For the model in Question 1, calculate the cost increase from \(q=10\) to \(q=20\).
- Why does knowing marginal cost without any baseline fail to determine one unique total-cost function?
- A piecewise cost function is continuous at a regime boundary, but its left- and right-hand derivatives are 7 and 11. Does its derivative exist at the boundary? Explain.
- What are the units of \(\int_{q_0}^{q_1}m(x)\,dx\) when cost is measured in dollars and production in items?
