Introduction
A firm planning production needs more than the total cost of meeting one particular output target. It may also need to know how much its least attainable cost will change if the target rises slightly. That rate of change is the marginal cost of increasing required output.
In this tutorial, output is a real number \(q\) in an operating range, and \(C(q)\) is the minimum cost of meeting that output requirement. When \(C\) is differentiable, its derivative \(C'(q)\) measures the local rate at which minimum cost changes as required output changes. The derivative is a rate, not generally the exact cost of a finite increase.
The Minimum-Cost Function
Let \(X(q)\) be the set of production plans that can meet a required output \(q\), and let \(c(\mathbf{x})\) be the cost of plan \(\mathbf{x}\). When a least-cost plan exists, define the minimum-cost function by minimizing the cost over all feasible plans:
The function \(C\) records the least cost for each output requirement. In an application, its formula may already have been obtained from a production model. The question here is how to differentiate that formula and interpret the result.
If cost is measured in dollars and output in units, then \(C'(q)\) has units of dollars per unit. More generally, its units are the units of cost divided by the units of output. This unit check is useful: \(C(q)\) is a total cost, whereas \(C'(q)\) is a rate of change in that total.
Why the Derivative Represents Marginal Cost
The derivative is defined through the change in minimum cost divided by the corresponding change in required output. For a nonzero increment \(h\), the difference quotient is
It is the average rate of cost change over the interval from \(q\) to \(q+h\). If the limit exists as \(h\) approaches zero, that average rate approaches the derivative:
Thus, differentiability means that for small \(h\), the cost change is approximately \(C'(q)h\). More precisely, the error in that approximation, divided by \(h\), tends to zero as \(h\) tends to zero. This is why the derivative describes the local marginal cost, while a finite change must be calculated from \(C(q+h)-C(q)\) if an exact amount is needed.
Worked Example: A Quadratic Minimum-Cost Function
Worked Example: A Quadratic Minimum-Cost Function
Suppose \(q\) is measured in units produced per week, and the minimum weekly cost, in dollars, is \(C(q)=500+12q+0.02q^2\). Find the marginal cost at \(q=100\), and compare it with the exact cost increase from raising output from 100 to 101 units per week.
Differentiate each term with respect to \(q\):
At \(q=100\), the marginal cost is
The marginal cost is therefore 16 dollars per additional unit at this output level. To find the exact increase for one more unit, calculate both costs:
The exact cost increase is \(1916.02-1900=16.02\) dollars per week. It differs slightly from the derivative value of 16 because one unit is a finite increase, not an infinitesimal one.
As a check on the derivative, use the difference quotient:
Taking \(h\) to zero gives 16, agreeing with \(C'(100)\). The exact one-unit increase also agrees with the quotient at \(h=1\): \(16+0.02(1)=16.02\).
A Monotonicity Check
There is a basic consistency check for a minimum-cost function. Suppose a larger required output means that every plan feasible for the larger requirement is also feasible for the smaller one. In symbols, if \(q_2>q_1\), suppose \(X(q_2)\subseteq X(q_1)\). Then the least cost cannot decrease as the requirement increases.
To see why, every plan in \(X(q_2)\) is also in \(X(q_1)\). The plans available for the larger requirement are therefore a subset of the plans available for the smaller requirement. Minimizing over fewer choices cannot produce a lower minimum than minimizing over the larger set of choices.
Where this nondecreasing function is differentiable, its derivative cannot be negative. For \(h>0\), the difference quotient \([C(q+h)-C(q)]/h\) is nonnegative. For \(h<0\), both the cost difference and \(h\) are nonpositive, so the quotient is again nonnegative. If the two-sided limit exists, it follows that \(C'(q)\ge 0\). This conclusion relies on nested feasible sets and a consistent cost function; if the model changes as output changes, the check need not apply.
Worked Example: A Cost Function with a Square Root
Worked Example: A Cost Function with a Square Root
Suppose \(C(q)=30q+120\sqrt{q}\) dollars for \(q>0\), where \(q\) is the required output in units. Find the marginal cost at \(q=16\). Verify the derivative using the difference quotient.
Differentiate the two terms. Since \(\frac{d}{dq}\sqrt{q}=\frac{1}{2\sqrt{q}}\) for \(q>0\),
At \(q=16\), \(\sqrt{16}=4\), so
The marginal cost is 45 dollars per additional unit at \(q=16\). To check directly, for \(h\ne 0\) near zero,
Rationalize the square-root difference using \((\sqrt{16+h}-4)(\sqrt{16+h}+4)=h\):
As \(h\) tends to zero, the denominator tends to \(4+4=8\). Therefore the quotient tends to \(30+120/8=30+15=45\), confirming the derivative calculation. The function is differentiable here because \(16\) is inside its domain \(q>0\).
When There Is No Single Marginal Cost
A minimum-cost function can have a kink. At such an output level, the rate of cost increase from below may differ from the rate from above. The ordinary derivative then does not exist, even though the minimum cost itself is continuous.
Worked Example: A Threshold with Different One-Sided Rates
Consider the cost function, in dollars, given by \(C(q)=10q+q^2\) for \(0\le q\le 10\), and \(C(q)=200+45(q-10)+2(q-10)^2\) for \(q\ge 10\). Find the marginal cost immediately to the left and right of \(q=10\), and determine whether \(C'(10)\) exists.
At the threshold, the first formula gives \(C(10)=10(10)+10^2=100+100=200\). The second formula also gives \(200+45(0)+2(0)^2=200\), so the cost is continuous there.
For \(q<10\), differentiating the first formula gives \(C'(q)=10+2q\), whose limit as \(q\) approaches 10 from the left is \(10+2(10)=30\). For \(q>10\), differentiating the second formula gives \(C'(q)=45+4(q-10)\), whose limit as \(q\) approaches 10 from the right is \(45+4(0)=45\).
The difference quotients at the threshold show the same result. If \(h<0\), then \(10+h\le 10\), and
As \(h\) approaches zero from the left, this tends to 30. If \(h>0\), then \(10+h\ge 10\), and
As \(h\) approaches zero from the right, this tends to 45. The two one-sided limits are unequal, so \(C'(10)\) does not exist. It is accurate to report the one-sided marginal costs, but not to assign a single derivative at the threshold.
Common Mistakes and Practical Interpretation
Confusing marginal cost with average cost. In the quadratic example, the average cost at \(q=100\) is \(C(100)/100=1900/100=19\) dollars per unit. That is not the marginal cost, which is \(C'(100)=16\) dollars per additional unit. Average cost divides total cost by total output; marginal cost differentiates total cost with respect to output.
Treating a derivative as an exact finite change. In that same example, the derivative at 100 predicts an increase of about \(16(1)=16\) dollars for one extra unit. The exact increase is \(16.02\) dollars. For larger changes, use \(C(q+h)-C(q)\) when an exact total change is needed. The derivative is a local rate, and its approximation is most reliable for small \(h\).
Ignoring the domain or a threshold. Before differentiating, check the output range where the formula applies. A formula may be differentiable for positive output but not at zero, or a piecewise cost model may have different rates on either side of a threshold. At a boundary, use an appropriate one-sided derivative if it exists; at a kink, report the distinct one-sided rates rather than a nonexistent ordinary derivative.
Marginal cost helps compare nearby production targets and estimate the cost consequences of a small output adjustment. Its interpretation depends on the units, the output range, and the assumptions used to define the minimum-cost function. A negative derivative under nested feasible sets is a reason to check the model or the calculation, since it conflicts with the monotonicity result above.
Check Your Understanding
Answer each question using the definitions and methods in this tutorial.
- If \(C(q)=80+9q+0.05q^2\), what is the marginal cost at \(q=40\), and what are its units if \(q\) is measured in units produced and \(C\) in dollars?
- In your own words, why is \(C'(q)\) usually an approximation to the cost of a finite output increase rather than the exact cost increase?
- For \(C(q)=15q+50\sqrt{q}\), find the marginal cost at \(q=25\). State the domain condition needed for the square-root derivative.
- How does average cost differ from marginal cost?
- At a threshold, the left-hand difference quotient tends to 24 and the right-hand difference quotient tends to 31. Does the ordinary derivative exist there? What can you report instead?