Why Derive Average and Marginal Cost?
A long-run total cost function describes the minimum cost of producing each possible output when the firm can choose its inputs. Two useful measures follow from it. Average cost reports cost per unit at a given output; marginal cost reports how total cost changes as output increases. Their relationship tells us whether average cost is rising or falling, a distinction that matters when interpreting a firm’s cost structure.
The formulas in this tutorial assume that input prices are fixed and that the long-run total cost function is differentiable at the positive output being considered. Output \(q\) is measured in units, and total cost \(C(q)\) is measured in dollars. A derivative describes a local rate of change; it is not automatically the exact cost of producing one additional whole unit.
Deriving the Relationship Between Average and Marginal Cost
Average cost is a ratio: divide total cost by output. Differentiating this ratio reveals how its slope depends on marginal cost. The derivation requires \(q>0\) and a differentiable \(C(q)\).
By definition, \(AC(q)=C(q)/q\).
The derivative of \(C(q)/q\) is \(\big(qC'(q)-C(q)\big)/q^2\).
Since \(C'(q)=MC(q)\) and \(C(q)/q=AC(q)\), the numerator becomes \(qMC(q)-qAC(q)\).
Dividing the numerator by \(q^2\) gives \(AC'(q)=\big(MC(q)-AC(q)\big)/q\).
Because \(q\) is positive, the denominator does not change the sign. Therefore, if marginal cost is below average cost, average cost is falling; if marginal cost is above average cost, average cost is rising. When the two are equal, average cost has a horizontal tangent. Equality alone identifies a stationary point; to establish a minimum, check how the slope changes around it or otherwise verify that average cost is minimized there.
Worked Applications
Worked Example: Derive Both Measures and Interpret Their Slopes
A workshop has long-run total cost \(C(q)=4q+q^2\), where \(q\) is the number of custom cases produced and \(C(q)\) is in dollars. Derive average and marginal cost, then evaluate them at \(q=6\) and determine whether average cost is rising or falling there.
Method. Divide total cost by output to obtain average cost, differentiate total cost to obtain marginal cost, and compare the two measures. This avoids treating average cost itself as marginal cost.
For \(q>0\), average cost is
Marginal cost is the derivative of total cost:
At \(q=6\), total cost is \(C(6)=4(6)+6^2=24+36=60\) dollars. Thus \(AC(6)=60/6=10\) dollars per case. Marginal cost is \(MC(6)=4+2(6)=16\) dollars per case. Since \(16>10\), average cost is rising at this output. Directly, \(AC'(q)=1\), so \(AC'(6)=1>0\), which confirms the conclusion.
Check. The relationship gives \(AC'(6)=(16-10)/6=1\), matching the direct derivative. The common mistake this method avoids is reporting \(C(6)=60\) as average cost; it is total cost, while average cost is \(60/6=10\) dollars per case.
Worked Example: Marginal Cost Below Average Cost
A small producer’s long-run total cost is \(C(q)=3q+12\sqrt{q}\) dollars for \(q>0\). Find average and marginal cost at \(q=16\), and determine the direction in which average cost is changing.
Method. Simplify the average-cost expression and differentiate total cost with respect to output. The derivative of \(\sqrt{q}\) is \(1/(2\sqrt{q})\), so the output units are not treated as if the square-root term were linear.
Average cost is
Marginal cost is
At \(q=16\), \(\sqrt{16}=4\). Total cost is \(C(16)=3(16)+12(4)=48+48=96\) dollars, so \(AC(16)=96/16=6\) dollars per unit. Marginal cost is \(MC(16)=3+6/4=4.5\) dollars per unit. Because \(4.5<6\), average cost is falling.
The derivative verifies this result: \(AC'(16)=(4.5-6)/16=-1.5/16=-0.09375\). The negative sign means average cost decreases as output increases locally at \(q=16\). The cost function is differentiable for \(q>0\); this calculation does not use a derivative at zero, where the square-root derivative is not finite.
Check. The formula gives \((MC-AC)/q=(4.5-6)/16=-0.09375\), as does differentiating \(3+12/\sqrt{q}\) and evaluating at 16. A common error is to conclude that average cost must rise because total cost rises with output. Total cost can rise while cost per unit falls.
Worked Example: Locate and Verify a Minimum of Average Cost
A firm’s long-run total cost for a specialized component is \(C(q)=q^3-9q^2+30q\) dollars. Find the output at which average cost is minimized, and compare average and marginal cost at outputs of 3, 4.5, and 6 units.
Method. Derive both cost measures, solve \(MC=AC\) for positive output, and check the sign of the average-cost slope on either side. This checks that the stationary point is a minimum rather than assuming every intersection is one.
For \(q>0\),
Differentiating total cost gives
Equating the two measures gives \(3q^2-18q+30=q^2-9q+30\), or \(2q^2-9q=0\). Factoring yields \(q(2q-9)=0\). Since the analysis concerns positive output, the relevant solution is \(q=4.5\). At this output, \(AC(4.5)=4.5^2-9(4.5)+30=20.25-40.5+30=9.75\) dollars per unit. Also, \(MC(4.5)=3(20.25)-18(4.5)+30=60.75-81+30=9.75\) dollars per unit.
To verify that average cost is minimized there, use the slope relationship:
At \(q=3\), \(AC(3)=9-27+30=12\), while \(MC(3)=27-54+30=3\). Marginal cost is below average cost, and \(AC'(3)=6-9=-3\), so average cost is falling. At \(q=6\), \(AC(6)=36-54+30=12\), while \(MC(6)=108-108+30=30\). Marginal cost is above average cost, and \(AC'(6)=12-9=3\), so average cost is rising. The slope changes from negative to positive at \(q=4.5\); average cost is minimized there.
Check. At the minimum, both cost measures equal \(9.75\) dollars per unit, and the slope is zero. The common mistake of accepting \(q=0\) from the factored equation is avoided by noting that \(AC=C/q\) is defined here only for positive output.
Using the Relationship Correctly
The derivative relationship provides a compact way to interpret a cost function, but it depends on comparing the correct quantities at the same output. Average cost is the cost per unit at that output. Marginal cost is the derivative of total cost there. Comparing marginal cost with total cost, or comparing the two measures at different outputs, does not establish whether average cost is rising or falling.
A second distinction is between an exact finite change and a marginal change. If output increases from \(q\) to \(q+1\), the exact change in total cost is \(C(q+1)-C(q)\). Marginal cost \(C'(q)\) is the instantaneous rate of change at \(q\); it can approximate that one-unit change when output is treated continuously, but need not equal it. The derivative formula applies where \(C\) is differentiable. At a kink or other point where the derivative does not exist, marginal cost in this sense is not defined at that point, although one-sided changes may still be examined.
These definitions and relationships are aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus. For exam calculations, show the expressions for both measures, substitute the specified output, keep units consistent, and state whether the comparison implies rising, falling, or stationary average cost.
Check Your Understanding
For each problem, derive or use the relevant cost measures before checking the final answer.
- A producer has \(C(q)=5q+q^2\). Find \(AC\) and \(MC\) at \(q=4\), and state whether average cost is rising or falling. (Answer: \(AC=9\), \(MC=13\) dollars per unit; average cost is rising.)
- A maker has \(C(q)=2q+18\sqrt{q}\). At \(q=9\), calculate average and marginal cost and determine the direction of the average-cost slope. (Answer: \(AC=8\), \(MC=5\) dollars per unit; average cost is falling.)
- For \(C(q)=q^3-8q^2+24q\), find the positive output where \(MC=AC\), and verify whether average cost has a minimum there. (Answer: \(q=4\); \(AC=MC=8\) dollars per unit, and the slope changes from negative to positive, so average cost is minimized.)
- A firm’s cost function is \(C(q)=q^2+6q\). At \(q=3\), calculate total, average, and marginal cost, and state whether average cost is rising or falling. (Answer: total cost \(27\) dollars, average cost \(9\) dollars per unit, marginal cost \(12\) dollars per unit; average cost is rising.)
- A supplier’s long-run cost is \(C(q)=0.5q^3-6q^2+30q\). Find the positive output where average cost is stationary, calculate both cost measures there, and verify the type of stationary point. (Answer: \(q=6\); \(AC=MC=12\) dollars per unit, and average cost is minimized.)