Deriving Marginal Revenue From Inverse Demand

Why Derive Marginal Revenue?

A demand schedule gives prices and quantities at selected points. For a firm choosing among such points, differences in total revenue help describe how revenue changes as sales change. When demand is instead represented by a differentiable inverse-demand curve, calculus gives the local rate of change directly. That rate is marginal revenue.

The distinction matters because increasing sales usually requires lowering the price charged for every unit sold, not just the additional unit. Marginal revenue therefore reflects both the revenue from the extra output and the effect of the price change on existing sales. The derivation below is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Definition: Let $Q$ be the quantity sold and let $P(Q)$ be the inverse-demand curve: the price per unit buyers will pay when quantity is $Q$. If the seller charges one price for every unit, total revenue is $TR(Q)=Q P(Q)$. Where $TR$ is differentiable, marginal revenue is the derivative of total revenue with respect to quantity: $$ MR(Q)=\frac{dTR(Q)}{dQ}. $$

Deriving the Marginal-Revenue Formula

Assume that quantity is a continuous choice, that the seller charges the same price for each unit, and that inverse demand $P(Q)$ is differentiable at the quantity being considered. Total revenue is quantity times price. Since both factors can change when quantity changes, apply the product rule:

\begin{aligned} TR(Q)&=Q P(Q),\\ MR(Q)&=\frac{d}{dQ}\big[Q P(Q)\big]\\ &=P(Q)+Q P'(Q). \end{aligned}

Here $P'(Q)$ is the slope of inverse demand: the local change in price per unit when quantity increases by one unit. The first term, $P(Q)$, is the revenue associated with one additional unit at the current price. The second term, $Q P'(Q)$, accounts for the price change on the quantity already being sold. For a downward-sloping inverse-demand curve, $P'(Q)<0$, so this second term reduces marginal revenue.

Key result: For differentiable inverse demand and single-price sales, marginal revenue at quantity $Q$ is $$ MR(Q)=P(Q)+Q P'(Q). $$ If $Q>0$ and demand slopes downward, then $Q P'(Q)<0$, so marginal revenue is below the current price. If inverse demand is locally flat, $P'(Q)=0$, and marginal revenue equals price at that quantity.

This is a local result: it measures the instantaneous rate at which revenue changes at a particular quantity. It is not generally the exact revenue change for a whole-unit increase or any other finite increase. For a finite change from $Q$ to $Q+\Delta Q$, the exact change is $TR(Q+\Delta Q)-TR(Q)$. The derivative approximates that change by $MR(Q)\Delta Q$ when $\Delta Q$ is sufficiently small.

Write total revenue.
Multiply quantity by the price given by inverse demand: $TR(Q)=QP(Q)$.

Differentiate both factors correctly.
Use the product rule to obtain $MR(Q)=P(Q)+QP'(Q)$.

Evaluate the price and slope.
Substitute the specified quantity into both $P(Q)$ and $P'(Q)$, then calculate marginal revenue.

Check the units and interpretation.
State whether the result is dollars per unit, or another rate consistent with how price and quantity are measured.

Worked Applications

Each example uses the same derived rule, but the form of inverse demand and the units differ. In each case, the slope must be evaluated at the stated quantity; using the price alone would omit the effect of a price change on existing sales.

Worked Example: Marginal Revenue on a Curved Demand Curve

A specialty supplier faces inverse demand $P(Q)=180-0.02Q^2$, where $Q$ is measured in individual units and $P$ is in dollars per unit. At $Q=50$, find price, total revenue, and marginal revenue. Also calculate the exact revenue change if sales rise from 50 to 51 units, and compare it with the marginal-revenue approximation.

Method. Differentiate total revenue using the product rule. This avoids the mistake of treating the current price as marginal revenue.

First calculate the price and total revenue at 50 units:

\begin{aligned} P(50)&=180-0.02(50)^2=180-50=130\text{ dollars per unit},\\ TR(50)&=50(130)=6{,}500\text{ dollars}. \end{aligned}

Differentiate inverse demand and evaluate its slope:

P'(Q)=-0.04Q,\qquad P'(50)=-0.04(50)=-2.

The slope is $-2$ dollars per unit of quantity for each additional unit sold, evaluated locally at 50. Thus:

MR(50)=P(50)+50P'(50)=130+50(-2)=130-100=30\text{ dollars per additional unit}.

For the exact one-unit change, calculate revenue at 51. The price is $P(51)=180-0.02(51)^2=180-52.02=127.98$ dollars per unit. Therefore:

TR(51)=51(127.98)=6{,}526.98\text{ dollars}, \qquad TR(51)-TR(50)=6{,}526.98-6{,}500=26.98\text{ dollars}.

The derivative gives a local estimate of 30 dollars for a one-unit increase, while the exact change is 26.98 dollars. They differ because the slope of total revenue changes over the finite interval from 50 to 51. The calculation checks: marginal revenue is below the current price of 130 dollars because inverse demand slopes downward.

Worked Example: Marginal-Revenue Units When Quantity Is in Thousands

A hypothetical distributor’s inverse demand is $P(Q)=135-0.01Q^2$, where $Q$ is measured in thousands of packages and $P$ is dollars per package. Total revenue is consequently measured in thousands of dollars. At $Q=50$, find price, total revenue, and marginal revenue. Then estimate the revenue change for an increase of one unit of $Q$, and check it against the exact change.

Method. Apply $MR=P+QP'$, keeping track of the quantity and revenue scales. The common unit trap is to divide by 1,000 twice when converting the derivative.

At $Q=50$ thousand packages:

\begin{aligned} P(50)&=135-0.01(50)^2=135-25=110\text{ dollars per package},\\ TR(50)&=50(110)=5{,}500\text{ thousand dollars}. \end{aligned}

Differentiate the inverse-demand function:

P'(Q)=-0.02Q,\qquad P'(50)=-0.02(50)=-1.

Using the derivative formula:

$$ MR(50)=110+50(-1)=60. $$

The units are thousand dollars per thousand packages. These scale factors cancel numerically: this is 60 dollars per package at the margin. Equivalently, a local increase of one thousand packages corresponds to approximately 60 thousand dollars of additional revenue. It is not 0.06 dollars per package.

Check the finite increase of one unit of $Q$, which means an increase of one thousand packages. At $Q=51$, price is $135-0.01(51)^2=135-26.01=108.99$ dollars per package. Revenue is $51(108.99)=5{,}558.49$ thousand dollars, so the exact change is $5{,}558.49-5{,}500=58.49$ thousand dollars. This is close to, but not exactly, the local estimate of 60 thousand dollars. The units and the direction of the change are consistent with downward-sloping demand.

Worked Example: Marginal Revenue From Rational Inverse Demand

A hypothetical seller faces inverse demand $P(Q)=240/(1+Q/20)$, with $Q$ in units and $P$ in dollars per unit. At $Q=20$, calculate the slope of inverse demand, total revenue, and marginal revenue. Then find the exact revenue change for an increase to 21 units.

Method. Rewrite the inverse-demand function in a form that is straightforward to differentiate, then compare the derivative with a finite revenue difference. The common mistake is to use the price at 20 as the revenue from one extra unit without accounting for the price reduction on earlier units.

Rewrite and evaluate price:

P(Q)=\frac{4{,}800}{Q+20}, \qquad P(20)=\frac{4{,}800}{40}=120\text{ dollars per unit}.

Total revenue at 20 is $20(120)=2{,}400$ dollars. Differentiate the rewritten demand curve:

P'(Q)=-\frac{4{,}800}{(Q+20)^2}, \qquad P'(20)=-\frac{4{,}800}{40^2}=-3.

Thus:

MR(20)=P(20)+20P'(20)=120+20(-3)=60\text{ dollars per additional unit}.

At 21 units, price is $4{,}800/41$ dollars per unit, and revenue is $21(4{,}800/41)=100{,}800/41$ dollars. The exact revenue change is:

\frac{100{,}800}{41}-2{,}400 =\frac{100{,}800-98{,}400}{41} =\frac{2{,}400}{41} \approx 58.54\text{ dollars}.

The exact change, rounded to two decimal places, is 58.54 dollars, compared with marginal revenue of 60 dollars. The derivative is a local rate, not an exact prediction for a one-unit increase. The exact change is positive but smaller than the local estimate, which is consistent with marginal revenue declining over this interval.

Interpreting the Slope and Knowing the Limits

The term $QP'(Q)$ explains why the slope matters. At a positive quantity, a steeper downward slope makes $P'(Q)$ more negative; holding the current price and quantity fixed, that makes $QP'(Q)$ more negative and reduces marginal revenue. A flatter demand curve has a smaller price effect on existing sales. If the curve is locally horizontal, the price effect vanishes and marginal revenue equals price at that point.

This rule depends on the stated setting. It describes a seller that chooses quantity along a differentiable inverse-demand curve and charges one price per unit. If prices differ across units, revenue need not be $QP(Q)$, so this derivation does not automatically apply. At a kink or discontinuity where a derivative does not exist, a single derivative-based marginal-revenue value may also be unavailable; the change in revenue must instead be analyzed using the relevant one-sided or finite changes.

A useful check is to ask whether the sign and size make economic sense. For positive quantity and downward-sloping demand, marginal revenue must be below price. It can still be positive, zero, or negative: the formula’s sign depends on whether the revenue from the extra unit is larger than, equal to, or smaller than the revenue reduction associated with the price change on existing units.

Common Mistakes

Using price as marginal revenue. The formula contains both $P(Q)$ and $QP'(Q)$; dropping the slope term ignores the effect of a price change on existing sales.
Applying the product rule incorrectly. Differentiating $QP(Q)$ gives $P(Q)+QP'(Q)$, not just $QP'(Q)$ or $P(Q)P'(Q)$.
Using the slope without evaluating it at the specified quantity. For a curved demand function, $P'(Q)$ generally changes with $Q$.
Calling a finite difference the derivative. The exact change over a nonzero interval is $TR(Q+\Delta Q)-TR(Q)$; marginal revenue is the local derivative at $Q$.
Misconverting scaled units. If quantity is measured in thousands and revenue in thousands of dollars, a derivative stated in thousand dollars per thousand units has the same numerical value as dollars per unit. Do not divide by 1,000 again.

Key takeaway: Under single-price sales with differentiable inverse demand, derive total revenue as $QP(Q)$ and differentiate it to obtain $MR(Q)=P(Q)+QP'(Q)$. The slope term captures the revenue effect of changing the price on units already being sold.

Check Your Understanding

For each problem, calculate the requested quantity and keep the units attached to your answer. Treat marginal revenue as a local derivative, not automatically as an exact finite change.

A seller faces $P(Q)=96-0.5Q$, with price in dollars per unit and quantity in units. At $Q=40$, find price, total revenue, and marginal revenue. (Answer: $P=76$ dollars per unit, $TR=3{,}040$ dollars, and $MR=56$ dollars per additional unit.)
Inverse demand is $P(Q)=150-2\sqrt{Q}$, with price in dollars per unit. At $Q=36$, calculate $P'(Q)$ and marginal revenue. (Answer: $P'(36)=-1/6$, and $MR(36)=132$ dollars per additional unit.)
A hypothetical producer faces $P(Q)=210-0.015Q^2$, where $Q$ is thousands of units and price is dollars per unit. At $Q=40$, state total revenue in thousands of dollars and marginal revenue in both thousand dollars per thousand units and dollars per unit. (Answer: $TR=7{,}440$ thousand dollars; $MR=138$ thousand dollars per thousand units, equivalently 138 dollars per unit.)
A seller’s inverse demand is $P(Q)=7{,}500/(Q+25)$, with $Q$ in units and price in dollars per unit. At $Q=25$, calculate marginal revenue and the exact revenue change from 25 to 26 units, rounding the latter to cents. (Answer: $MR=75$ dollars per additional unit; the exact change is $2{,}500/34\approx 73.53$ dollars.)
At $Q=120$, inverse demand is locally differentiable, the price is 64 dollars per unit, and its slope is zero. Find marginal revenue and explain why it equals price. (Answer: $MR=64$ dollars per additional unit, because $QP'(Q)=120(0)=0$.)

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