Deriving Marginal Revenue From Total Revenue

Why Derive Marginal Revenue?

Total revenue records the money received from selling output. Marginal revenue describes how that total changes when output increases. The distinction matters because the additional revenue from a sale need not equal the price: if a seller must lower the price on all units to sell one more, the extra unit changes revenue on both the new unit and the units already sold.

The appropriate calculation depends on how output is represented. When output changes in whole units, compare total revenue at adjacent quantities. When output is modelled as continuous, marginal revenue is the derivative of total revenue with respect to output. These measures are related, but a one-unit finite change is not generally identical to the derivative at the starting quantity.

Definition: If \(R(q)\) is total revenue at output \(q\), discrete marginal revenue from increasing output from \(q\) to \(q+1\) is \(R(q+1)-R(q)\). For continuous output, marginal revenue is \(MR(q)=\frac{dR(q)}{dq}\), the instantaneous rate of change in total revenue with respect to output. If revenue is measured in dollars and output in units, marginal revenue is measured in dollars per unit.

Discrete Output: Revenue Change Between Adjacent Quantities

For integer output, each additional unit is a finite change. If \(p(q)\) is the price per unit when \(q\) units are sold under a single-price schedule, total revenue is \(R(q)=q\,p(q)\). The marginal revenue from producing and selling the next unit is therefore

\Delta R=R(q+1)-R(q)=(q+1)p(q+1)-qp(q).

This is an exact difference, not an approximation. It also shows why marginal revenue may differ from the next unit’s selling price. Rearranging gives

\Delta R=p(q+1)+q\bigl[p(q+1)-p(q)\bigr].

The first term is the revenue from the additional unit at the new price. The second term is the revenue change on the \(q\) units that could now sell only at that new price. If price falls when output rises, that second term is negative.

Worked Example: Finding the Revenue From the Ninth Unit

A hypothetical seller faces the single-price schedule \(p(q)=70-2q\) dollars per unit, where \(q\) is an integer number of units. Find total revenue at outputs of 8 and 9, then find the discrete marginal revenue from increasing output from 8 to 9.

Method. Calculate total revenue at both adjacent output levels and take the difference. This avoids treating the new unit’s price as the full revenue gain.

At \(q=8\), the price is \(p(8)=70-2(8)=54\) dollars. Thus,

R(8)=8(54)=432\text{ dollars}.

At \(q=9\), the price is \(p(9)=70-2(9)=52\) dollars. Thus,

R(9)=9(52)=468\text{ dollars}.

The change in revenue is

R(9)-R(8)=468-432=36\text{ dollars}.

As a check, the ninth unit brings in \(52\) dollars at the new price, while the price reduction of \(2\) dollars applies to the 8 earlier units: \(52-8(2)=36\). The result is an exact discrete marginal revenue of \(36\) dollars for this one-unit increase.

Continuous Output: Marginal Revenue as a Derivative

In a continuous model, output \(q\) can vary over an interval rather than only by whole units. Assume total revenue is a differentiable function \(R(q)\) on the output range being considered. The derivative \(dR/dq\) gives the slope of the total-revenue curve at \(q\): it measures the instantaneous rate at which revenue changes for a small change in output.

If a single price \(p(q)\) depends on the quantity sold, total revenue is \(R(q)=q\,p(q)\). Provided \(p(q)\) is differentiable, applying the product rule gives

MR(q)=\frac{dR}{dq} =\frac{d}{dq}\bigl[q\,p(q)\bigr] =p(q)+q\frac{dp(q)}{dq}.

The formula has a direct interpretation. The term \(p(q)\) is the price component of the revenue change. The term \(q\,dp(q)/dq\) captures how a price change affects revenue on the existing \(q\) units. With a downward-sloping price schedule, \(dp/dq<0\), so marginal revenue is below price for positive output.

Derivation: Total revenue is the product of quantity and price, \(R(q)=q\,p(q)\). For a small output increase \(dq\), the product changes by approximately \(p(q)\,dq+q\,dp\): the new quantity is valued at the current price, and the price change applies to the current quantity. Dividing by \(dq\) and taking the limit as \(dq\) approaches zero gives \(dR/dq=p(q)+q\,dp/dq\), which is the product rule. This derivation requires a differentiable price function; for integer-only output, use the exact finite difference instead.

Worked Example: Deriving Marginal Revenue From Inverse Demand

A hypothetical seller faces a continuous inverse-demand function \(p(q)=96-1.2q\), with price in dollars per unit and output \(q\) measured in units. At \(q=25\), find total revenue, average revenue, and marginal revenue. Then calculate the exact revenue change from increasing output from 25 to 26 units.

Method. First form total revenue from price times quantity. Differentiate that function for continuous marginal revenue, then separately calculate the adjacent-output difference. This keeps an instantaneous derivative distinct from a one-unit change.

At \(q=25\), price is

p(25)=96-1.2(25)=96-30=66\text{ dollars per unit}.

Total revenue is \(R(q)=q(96-1.2q)=96q-1.2q^2\), so

R(25)=96(25)-1.2(25^2)=2{,}400-750=1{,}650\text{ dollars}.

Average revenue is total revenue divided by output: \(1{,}650/25=66\) dollars per unit, equal here to the single price. Differentiating total revenue gives

MR(q)=96-2.4q,\qquad MR(25)=96-2.4(25)=96-60=36\text{ dollars per unit}.

For the exact one-unit change, at \(q=26\) the price is \(96-1.2(26)=64.8\) dollars, and revenue is \(R(26)=26(64.8)=1{,}684.8\) dollars. Therefore,

R(26)-R(25)=1{,}684.8-1{,}650=34.8\text{ dollars}.

The finite change is \(34.8\) dollars, whereas the derivative at \(q=25\) is \(36\) dollars per unit. They are not contradictory: the derivative is the slope at the starting quantity, while the finite change covers the whole interval from 25 to 26. As a check, price has fallen by \(1.2\) dollars on the 25 existing units, so the next unit’s \(64.8\) dollars of revenue is offset by \(25(1.2)=30\) dollars: \(64.8-30=34.8\).

Starting With a Total-Revenue Function

Sometimes a problem gives \(R(q)\) directly rather than giving a price schedule. For a differentiable total-revenue function, differentiate with respect to output. If the problem asks for the revenue change over a whole unit, evaluate the function at both endpoints and subtract. These answer different questions, so follow the wording and the output model carefully.

Worked Example: Comparing the Slope With a One-Unit Change

Suppose a hypothetical seller’s continuous total-revenue function, in dollars, is \(R(q)=400q-4q^2+0.02q^3\), where \(q\) is output in units. Find marginal revenue at \(q=10\), then calculate the exact change in total revenue from 10 to 11 units.

Method. Differentiate for the instantaneous marginal revenue and evaluate the total-revenue function at both quantities for the finite change. The function is differentiable, so the derivative is defined; the finite difference remains the appropriate calculation for the specified one-unit interval.

Differentiate each term:

MR(q)=\frac{dR}{dq}=400-8q+0.06q^2.

At \(q=10\),

MR(10)=400-8(10)+0.06(10^2)=400-80+6=326\text{ dollars per unit}.

Total revenue at 10 units is

R(10)=400(10)-4(10^2)+0.02(10^3)=4{,}000-400+20=3{,}620\text{ dollars}.

At 11 units,

R(11)=400(11)-4(11^2)+0.02(11^3) =4{,}400-484+26.62=3{,}942.62\text{ dollars}.

Thus, the one-unit increase changes revenue by \(3{,}942.62-3{,}620=322.62\) dollars. The derivative at 10 is \(326\) dollars per unit, not the exact revenue change over the entire interval. The finite-change calculation checks against the two stated total revenues.

Common Mistakes and Checks

A frequent error is to identify marginal revenue with price. In the continuous inverse-demand example, price at \(q=25\) is \(66\) dollars. Calling \(66\) dollars marginal revenue is wrong: the price falls as output rises, so the additional output also reduces revenue on units already sold. The correct derivative is \(96-2.4(25)=36\) dollars per unit.

Another error is to substitute a finite change for a derivative, or a derivative for an exact one-unit change, without checking what the question asks. In that same example, \(36\) is the instantaneous slope at 25, while \(34.8\) is the exact revenue increase from 25 to 26. Both calculations are valid for their respective meanings.

Before finalizing a result, check that the units make sense, that total revenue has been calculated as price times quantity when a single-price schedule is given, and that a discrete change uses two adjacent revenue values. For a constant price \(p\), \(R(q)=pq\) and \(MR=p\); this is a useful check on the derivative method when price does not vary with output.

Key takeaway: For whole-unit output, marginal revenue is the exact difference \(R(q+1)-R(q)\). For a differentiable continuous total-revenue function, it is the derivative \(dR/dq\). If total revenue is \(q\,p(q)\), the derivative is \(p(q)+q\,dp/dq\), not generally just the price.

Check Your Understanding

For each problem, identify whether it asks for a discrete change or a derivative, and show the relevant revenue calculations.

A seller charges according to the integer-output schedule \(p(q)=52-2q\) dollars per unit. Find total revenue at 6 and 7 units and the discrete marginal revenue from the increase. (Answer: \(R(6)=240\) dollars, \(R(7)=266\) dollars, and discrete marginal revenue \(=26\) dollars.)
A continuous inverse-demand function is \(p(q)=84-1.5q\). Find total revenue, average revenue, and marginal revenue at \(q=12\). (Answer: total revenue \(=792\) dollars, average revenue \(=66\) dollars per unit, and marginal revenue \(=48\) dollars per unit.)
A continuous total-revenue function is \(R(q)=300q-3q^2+0.05q^3\), in dollars. Find marginal revenue at \(q=10\) and the exact revenue change from 10 to 11 units. (Answer: \(MR(10)=255\) dollars per unit; \(R(10)=2{,}750\) dollars; \(R(11)=3{,}003.55\) dollars; the change is \(253.55\) dollars.)
A seller’s inverse-demand schedule is \(p(q)=110-2q\) dollars per unit, and output is restricted to whole units. Find the exact revenue change from 20 to 21 units, then find the derivative-based marginal revenue at \(q=20\) using the continuous formula. (Answer: the exact change is \(28\) dollars; the derivative-based marginal revenue is \(30\) dollars per unit.)
A continuous seller faces constant price \(32\) dollars per unit. Write the total-revenue function and find marginal revenue at \(q=12\). Explain the relationship between price and marginal revenue in this case. (Answer: \(R(q)=32q\), and \(MR(12)=32\) dollars per unit, equal to the constant price.)

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