Why Average Cost Is a Different Objective
Production decisions often involve two different questions. One asks how much it costs to produce a quantity of output. The other asks how much it costs, on average, to produce each unit. These are not the same objective: the output level with the lowest total cost need not have the lowest cost per unit.
Let \(q\) denote the quantity produced, and let \(C(q)\) be the total cost of producing \(q\). We consider positive output, because average cost per unit is not defined at \(q=0\). If a production model permits outputs only in a specified range, that range is the feasible production range; optimization must take place within it.
The feasible range matters as much as the formula. A derivative can identify possible interior candidates, but it does not decide whether an output is allowed, whether a boundary is included, or which candidate gives the global minimum. Those checks are part of the optimization.
How the Derivative of Average Cost Works
Suppose \(C\) is differentiable at a positive output \(q\). Differentiating the quotient \(A(q)=C(q)/q\) gives
Because \(q^2>0\) for positive output, the sign of \(A'(q)\) is the sign of \(qC'(q)-C(q)\). Dividing that expression by \(q>0\) gives an especially useful form:
The derivative \(C'(q)\) is the marginal cost: it describes the instantaneous rate at which total cost changes as output changes. The quotient \(C(q)/q\) is average cost. Thus the derivative of average cost compares marginal cost with average cost.
The equality has a useful interpretation. If marginal cost is below average cost, then \(A'(q)<0\), so average cost is decreasing as output increases locally. If marginal cost is above average cost, then \(A'(q)>0\), so average cost is increasing locally. Equality marks a stationary point, not automatically a minimum; the feasible range and the behavior on either side still need to be checked.
Why Marginal Cost Equals Average Cost at an Interior Candidate
The condition follows directly from the quotient rule and the necessary derivative condition for an interior local minimum. In detail, assume \(q_*>0\), that \(C\) is differentiable near \(q_*\), and that \(q_*\) is an interior local minimum of \(A\). At a differentiable interior local minimum, the derivative is zero. Therefore
The denominator \(q_*^2\) is positive, so the numerator must be zero:
Since \(q_*>0\), divide by \(q_*\) to obtain
This proves the condition. It is a necessary condition for an interior minimum, not a complete test by itself: a stationary point could instead be a maximum or another type of stationary point. The derivative’s sign, other candidates, and feasible boundaries must still be considered.
A Practical Candidate Check
When the feasible production range is a closed interval \([a,b]\) with \(0<a<b\), and \(A\) is continuous on that interval and differentiable in its interior, a global minimum exists. To find it, check the endpoints and the interior critical points where \(A'(q)=0\). Any interior global minimum must be a critical point, while an endpoint minimum need not have derivative zero.
Record the allowed outputs, including whether each boundary is included. Exclude \(q=0\) from the average-cost formula.
Use \(A(q)=C(q)/q\), then solve \(A'(q)=0\) for feasible interior candidates.
Compare average cost at the interior candidates and at all included endpoints. For an excluded boundary, examine limiting behavior instead of treating it as an attained candidate.
On an open or half-open feasible range, the closed-interval existence guarantee does not apply automatically. A value approached near an excluded boundary may be an infimum without being a minimum. The definition of the feasible range therefore determines not only which outputs to compare, but also whether the lowest limiting value can actually be achieved.
Worked Examples
Worked Example: An Interior Minimum on a Closed Range
A producer has total cost \(C(q)=q^2+100\), with feasible output \(2\leq q\leq12\). Find the output that minimizes average cost.
First form average cost, keeping \(q>0\):
Differentiate and solve for interior critical points:
Output is positive, so the feasible solution is \(q=10\). The derivative is negative for \(2\leq q<10\) and positive for \(10<q\leq12\), since \(q^2<100\) on the first range and \(q^2>100\) on the second. Thus average cost decreases up to \(10\) and increases after it. Check the candidate and both endpoints:
Since \(20<61/3<52\), the minimum occurs at \(q=10\), with average cost \(20\) cost units per unit of output. Check using total cost: \(C(10)=10^2+100=200\), and \(C(10)/10=200/10=20\). Also \(C'(q)=2q\), so \(C'(10)=20=A(10)\), as the interior condition predicts.
Worked Example: A Minimum at the Feasible Endpoint
Suppose \(C(q)=q^2+4q+16\), and production is restricted to \(1\leq q\leq3\). Determine the output with minimum average cost.
Divide the total cost by positive output and differentiate:
For every feasible \(q\), \(q^2\leq9\), so \(16/q^2\geq16/9>1\). Hence \(A'(q)<0\) throughout the feasible interval: average cost decreases as output increases. There is no interior critical point, and the minimum is at the included right endpoint. Evaluate both endpoints:
Because \(37/3<21\), the minimizing output is \(q=3\). Check directly: \(C(3)=3^2+4(3)+16=9+12+16=37\), and \(C(3)/3=37/3\). Here \(C'(3)=2(3)+4=10\), which is not equal to \(A(3)=37/3\). There is no contradiction: the equality condition is for an interior differentiable minimum, whereas this minimum occurs at a feasible endpoint.
Worked Example: An Excluded Boundary and No Attained Minimum
Let \(C(q)=q^2+4q\) and suppose feasible outputs satisfy \(0<q\leq4\). Does average cost attain a minimum?
For every feasible output, divide by \(q>0\):
Average cost is strictly increasing, so smaller feasible outputs have smaller average cost. The boundary \(q=0\) is not feasible, and \(A(0)\) is not defined by the average-cost formula. As output approaches zero through positive values,
For every feasible \(q>0\), \(A(q)=q+4>4\). Thus no feasible output has average cost \(4\), although values can be made arbitrarily close to \(4\) by choosing positive output sufficiently close to zero. The infimum is \(4\), but there is no attained minimum. This conclusion respects the feasible range: an excluded boundary cannot be selected as the answer.
A Common Mistake: Minimizing Total Cost Instead
A frequent error is to minimize \(C(q)\) when the question asks for minimum average cost. In the first example, differentiating total cost gives \(C'(q)=2q\), which is positive for every feasible \(q\). Total cost therefore increases throughout \([2,12]\), so its minimum is at \(q=2\). But the average-cost calculations show \(A(2)=52\) and \(A(10)=20\), so producing \(10\) units gives lower average cost. The two objectives lead to different decisions.
A related algebra mistake is to differentiate \(C(q)/q\) as if its derivative were just \(C'(q)/q\). For the first example, that incorrect calculation would give \(2q/q=2\), falsely suggesting that average cost is always increasing. The quotient rule instead gives
At \(q=10\), this derivative is \(1-100/100=0\), consistent with the computed interior minimum. Always differentiate the full quotient, and always compare the resulting candidates within the feasible range.
Why the Boundary and Units Matter
Average cost is a per-unit quantity, so its units are cost per unit of output—not simply cost units. The units also help distinguish the objects in the calculation: \(C(q)\) is total cost, \(A(q)\) is cost per unit, and \(C'(q)\) is marginal cost per unit of additional output. In the interior condition \(C'(q)=A(q)\), the two sides therefore have matching units.
Boundary checks are equally important. A derivative equation identifies stationary outputs only where it applies; it does not include feasible endpoints automatically. If a boundary is excluded, a limiting value there may fail to be attained, as the third example demonstrates. A sound average-cost optimization therefore starts with the model and feasible outputs, not with differentiation alone.
Check Your Understanding
Use the definitions and checks in this tutorial to answer the following questions.
- If total cost is measured in dollars and output in units, what are the units of average cost?
- For positive \(q\), derive the formula for \(A'(q)\) from \(A(q)=C(q)/q\).
- What equality relates marginal cost and average cost at an interior differentiable minimum, and why does it not automatically apply at an endpoint?
- Why must an optimizer compare feasible endpoints as well as interior critical points on a closed production interval?
- How can the infimum of average cost fail to be a minimum when the feasible range excludes a boundary?
