Cost Change at a Particular Output
A business may want to estimate how its costs will change if it produces one more item. The answer depends on its current production level: the extra cost near a small output may differ from the extra cost near a large one. A cost function lets us describe that dependence.
The derivative gives a precise meaning to the local rate of change of cost with respect to output. When evaluated at a specified output, it is called marginal cost. It is a local rate, not the total cost of production and not necessarily the exact cost of making one additional item.
The difference quotient compares the cost at \(q_0+h\) with the cost at \(q_0\), then divides the cost change by the output change. For a finite \(h\), it is an average rate of change over that interval. Taking the limit as \(h\) approaches zero gives the instantaneous rate at the specified output.
In a mathematical model, output is often allowed to vary continuously even though actual production occurs in whole items. The derivative describes the model’s local behavior. It can estimate the cost change for a small increase in production, but it need not equal the exact change for an entire additional item.
Why the Derivative Represents Local Cost Change
The interpretation follows directly from the definition. Write the average cost change from \(q_0\) to \(q_0+h\) as a difference quotient. If that quotient approaches \(C'(q_0)\), then, for sufficiently small nonzero \(h\), it is close to \(C'(q_0)\). Multiplying by \(h\) shows that the cost change is close to the derivative times the output change.
For \(h\ne 0\), the average rate of cost change is \(\frac{C(q_0+h)-C(q_0)}{h}\).
If \(C'(q_0)\) exists, this quotient approaches \(C'(q_0)\) as \(h\to 0\). Equivalently, the quotient can be written as \(C'(q_0)+\varepsilon(h)\), where \(\varepsilon(h)\to 0\).
Multiplying by \(h\) gives \(C(q_0+h)-C(q_0)=C'(q_0)h+h\varepsilon(h)\). The remaining term, \(h\varepsilon(h)\), becomes negligible relative to \(h\) as \(h\to 0\).
This is why \(C'(q_0)h\) is a good local estimate of the cost change for a small output change \(h\). It is an approximation for small changes, not a claim that the exact cost change always equals \(C'(q_0)h\) for a finite change.
Worked Examples
Worked Example: Marginal Cost at 100 Items
Suppose \(q\) is output in items and \(C(q)=500+12q+0.02q^2\) is cost in dollars. Find and interpret the marginal cost at \(q_0=100\).
First calculate the cost at 100 items:
Now calculate the cost at \(100+h\) items:
There is an arithmetic issue in the last simplification if we combine terms carelessly: the constant term from the squared expression has already been included in \(C(100)=1900\). Expanding directly and grouping correctly gives:
Therefore, for \(h\ne 0\),
Taking the limit as \(h\to 0\) gives \(C'(100)=16\). The marginal cost is 16 dollars per item at an output of 100 items. In other words, the model’s cost is changing locally at about $16 for each additional item near that production level.
Check the interpretation against a one-item increase. The model gives \(C(101)=1900+16+0.02=1916.02\), so \(C(101)-C(100)=16.02\) dollars. This is close to, but not exactly, the marginal cost of $16 per item.
Worked Example: Marginal Cost at 25 Items
Let \(C(q)=100+3q+0.04q^2\), with output in items and cost in dollars. Find the marginal cost at 25 items, then compare it with the cost change for one item more and one item fewer.
At 25 items,
For an output change \(h\), expand the cost:
Thus,
The marginal cost at 25 items is 5 dollars per item. For a one-item increase, the exact change is \(C(26)-C(25)=5+0.04=5.04\) dollars. For one item fewer, calculate \(C(24)-C(25)\) by setting \(h=-1\): the change is \(5(-1)+0.04(-1)^2=-4.96\) dollars. Producing one fewer item reduces the modeled cost by $4.96.
Check: the quotient \(5+0.04h\) approaches 5 from either side as \(h\to 0\). The finite changes, $5.04 for an increase and \(-$4.96 for a decrease, are consistent with a local rate of $5 per item.
Worked Example: Converting Scaled Units
Suppose \(x\) measures output in hundreds of items and \(C(x)=20+3x+0.5x^2\) measures cost in thousands of dollars. Find the marginal cost when output is 400 items, and express it in dollars per item.
Since \(x\) is measured in hundreds of items, 400 items corresponds to \(x=4\). The cost at that output is
thousand dollars. For a change \(h\) in the \(x\)-coordinate,
It follows that
The derivative’s units are thousands of dollars per hundred items, because one unit of \(x\) represents 100 items and one unit of \(C\) represents $1,000. Convert the rate:
Check with a finite increase of one \(x\)-unit, which represents 100 items. The model gives \(C(5)=20+15+12.5=47.5\) thousand dollars, so the cost rises by \(47.5-40=7.5\) thousand dollars. That is an average increase of \(\$7{,}500/100=\$75\) per item across those 100 items, close to the local marginal cost of $70 per item.
Marginal Cost Is Not Total or Average Cost
The value \(C(q_0)\) is the total modeled cost at output \(q_0\). The marginal cost \(C'(q_0)\) is a rate of change at that output. These quantities answer different questions and generally have different units: total cost is measured in dollars, while marginal cost is measured in dollars per item.
Average cost also differs from marginal cost. At a positive output \(q\), average cost per item is \(C(q)/q\). It describes total cost divided by the number of items produced. Marginal cost describes how cost is changing near the current output. An average over all items produced so far does not, in general, tell us the local cost change near the next item.
A Common Pitfall: Treating a Finite Difference as the Derivative
For the first cost model, the one-item increase from 100 to 101 items costs \(C(101)-C(100)=\$16.02\). It would be incorrect to conclude that \(C'(100)=\$16.02\) just because the output increased by one item. That difference is an average rate over the interval from 100 to 101 items; the derivative is a limit of average rates as the interval becomes arbitrarily small.
The correct calculation uses a general change \(h\):
The one-item difference is useful as a nearby-output check, but it is not the definition of marginal cost. A finite increase may be a good approximation to the derivative when the increase is small; it is exact only in special cases, such as a cost function that is linear over the interval.
Using Marginal Cost Carefully
To interpret a marginal-cost value, identify both the output level and the units. For example, “marginal cost is 16” is incomplete unless the cost and output units are understood. A clear interpretation is: “At 100 items, the modeled cost is changing at approximately $16 per additional item.”
The estimate applies locally. It does not say that every item costs $16 to produce, that the total cost is $16, or that increasing production by any large number of items will raise cost by exactly $16 times that number. For a larger output change, the cost function itself or a suitable finite difference is needed to calculate the actual change.
Check Your Understanding
Use the definition of marginal cost and the role of units to answer these questions.
- If \(C(q)\) is measured in dollars and \(q\) in items, what are the units of \(C'(q_0)\)?
- For \(C(q)=80+4q+0.1q^2\), find \(C(10)\), form the difference quotient at \(q_0=10\), and calculate \(C'(10)\).
- For the cost function in question 2, what is the exact cost increase from 10 to 11 items? How does it compare with \(C'(10)\)?
- A model uses output in hundreds of items and cost in thousands of dollars. If its derivative is 3.5 in those model units, convert the marginal cost to dollars per item.
- Explain why a one-item cost difference is not, in general, the derivative at the starting output.
