Revenue and Its Marginal Rate
A business may want to know how revenue responds when it produces and sells a little more. The answer can depend on its current output level. Marginal revenue describes that local response: it is the rate at which total revenue changes with output at a specified level.
Let \(q\) denote output, measured in units of a product, and let \(R(q)\) denote the corresponding total revenue, measured in dollars. When \(R\) is differentiable at \(q\), its marginal revenue at that output is the derivative \(R'(q)\).
If output is measured in items and revenue in dollars, marginal revenue has units of dollars per item. The value \(MR(100)=12\), for example, means that near an output of 100 items, total revenue is changing at about 12 dollars for each additional item. It does not say that total revenue is 12 dollars.
The word marginal signals a rate at a particular output level. To compare marginal revenue across production levels, evaluate the marginal-revenue function at each level, or read the corresponding entries from a table. Then interpret each value with the same units and in the context of a small output change.
From a Rate to a Small Revenue Change
Suppose output increases from \(q\) to \(q+h\), where \(h\) is a small positive amount. The change in total revenue is \(R(q+h)-R(q)\). At a point where \(R\) is differentiable, this change can be estimated using the marginal revenue at the starting output:
The units provide a useful check: dollars per item multiplied by items gives dollars. This is a local estimate, not generally an exact formula for a finite increase. For a larger change in output, the marginal-revenue value may vary substantially between the starting and ending levels.
The approximation uses the marginal revenue at the starting level \(q\). It should not be confused with evaluating marginal revenue at \(q+h\), which describes the local rate near the ending level. Both rates can be useful, but they answer different local questions.
Why the Local Approximation Works
The derivative is defined as a limit of average rates of change. Where \(R\) is differentiable at \(q\),
This limit explains why multiplying marginal revenue by a small output change estimates the revenue change. Here is a precise version. Given any \(\varepsilon>0\), differentiability means that for all sufficiently small nonzero \(h\),
For a positive output increase \(h\), multiply this inequality by \(h\). The result is
Thus, for small enough \(h\), the actual revenue change differs from \(MR(q)h\) by less than \(\varepsilon h\). Since \(\varepsilon\) can be chosen as small as desired, the approximation becomes accurate relative to the size of the output change as \(h\) tends to zero. The differentiability condition matters: at a point where the derivative does not exist, this conclusion is not guaranteed by the derivative definition.
Worked Example: Comparing Two Levels from a Function
Worked Example: Comparing Two Levels from a Function
Suppose marginal revenue is given by \(MR(q)=80-0.4q\), in dollars per item, where \(q\) is the number of items produced and sold. Compare the local revenue response to an increase of 2 items at outputs of 50 items and 100 items.
First evaluate the marginal-revenue function at 50:
At 100 items:
Use each rate at its own starting output level to estimate the revenue change for 2 additional items:
The calculation checks by units: \((\text{dollars per item})(\text{items})=\text{dollars}\). The two function evaluations also check directly: substituting 50 gives \(80-20=60\), and substituting 100 gives \(80-40=40\). Under this model, the same small increase in output has a larger estimated revenue effect near 50 items than near 100 items.
Reading a Marginal-Revenue Table
A table supplies marginal-revenue values at listed output levels. Each entry is interpreted as a local rate at its row’s output. If the exact output is not listed, the table alone does not give an exact marginal-revenue value there. Any estimate between rows would require an additional assumption, such as a stated interpolation method.
Worked Example: Comparing Values in a Table
A company has the following marginal-revenue data. Output is measured in items and marginal revenue in dollars per item.
| Output \(q\) (items) | Marginal revenue \(MR(q)\) (dollars per item) |
|---|---|
| 0 | 12 |
| 20 | 9 |
| 40 | 6 |
| 60 | 3 |
Compare the estimated revenue changes for a 5-item increase starting at 20 items and starting at 60 items.
At 20 items, the table gives \(MR(20)=9\) dollars per item. Therefore,
At 60 items, the table gives \(MR(60)=3\) dollars per item. Therefore,
Both estimates pass the unit check, and each uses the table entry at the starting output level. The comparison is that the estimated revenue response to the same 5-item increase is larger near 20 items. The table does not establish the exact revenue change across either five-item interval; it supplies local rates at the listed starting levels.
Interpreting Positive, Zero, and Negative Values
The sign of marginal revenue indicates the direction of the local revenue response to a small positive output increase. If \(MR(q)>0\), the estimate \(MR(q)h\) is positive. If \(MR(q)=0\), the linear local estimate is zero. If \(MR(q)<0\), the estimate is negative. These are statements about the change in revenue near the specified output, not about whether total revenue itself is positive or negative.
Worked Example: Comparing Positive, Zero, and Negative Marginal Revenue
Suppose \(MR(q)=18-0.06q\) dollars per item. Compare the local responses to a 5-item output increase at \(q=200\), \(q=300\), and \(q=400\).
At 200 items:
The estimated revenue change is \(6(5)=30\) dollars. At 300 items:
so the linear local estimate for 5 more items is \(0(5)=0\) dollars. At 400 items:
so the estimated revenue change is \((-6)(5)=-30\) dollars. The calculations check by substitution, and each rate multiplied by 5 items has units of dollars.
The function is zero at \(q=300\), since solving \(18-0.06q=0\) gives \(0.06q=18\) and \(q=300\); substituting back gives \(18-0.06(300)=0\). Therefore this model’s marginal revenue is positive at 200 items, zero at 300, and negative at 400. In particular, negative marginal revenue at 400 means total revenue is locally decreasing as output increases from that level. It does not mean total revenue is negative.
Common Pitfall: Treating a Rate as Total Revenue
A frequent error is to read a marginal-revenue value as the total revenue, or to report the rate itself as the revenue change for any chosen increase. For example, if \(MR(50)=60\) dollars per item, saying “revenue increases by 60 dollars when output rises by 5 items” skips the multiplication by the output change.
Even the corrected value is a local estimate, not automatically an exact five-item revenue change. Another error is to compare rates without checking units: rates measured in dollars per item and thousands of dollars per thousand items may have different numerical representations. Convert to consistent units before comparing.
Using Marginal Revenue Responsibly
Marginal revenue is useful when the question concerns a small change near a specified production level. A careful comparison identifies the output level, reads or evaluates \(MR(q)\) there, checks the units, and then uses the rate to estimate the effect of a stated small output change. The sign indicates the estimated direction of that local effect.
The estimate should not be stretched beyond what the derivative says. A marginal-revenue value at one output level does not by itself determine the exact revenue change over a large interval. If output is restricted to whole items, a derivative-based model still describes a local rate; the exact change from one integer quantity to another would require the revenue values at both quantities.
Check Your Understanding
Answer each question using the output level and units stated.
- If \(MR(40)=7\) dollars per item, what is the estimated revenue change for a small increase of 3 items starting at 40 items?
- A table gives \(MR(100)=5\) dollars per item and \(MR(200)=2\) dollars per item. At which output level is the estimated revenue response larger for the same small positive increase? Explain.
- What does \(MR(q)=0\) say about the linear local estimate of revenue change for a small increase? Does it say that total revenue is zero?
- A marginal-revenue value is \(-4\) dollars per item. What is the estimated revenue change for a small increase of 2 items, and what does its sign mean?
- Why should \(MR(q)h\) generally be treated as an estimate rather than an exact revenue change over a finite interval?
