Linear Demand and Its Marginal Revenue Curve

From Linear Demand to Revenue

A linear inverse-demand curve describes the price buyers will pay for each quantity sold. Its shape makes it possible to derive total revenue and marginal revenue directly, and it reveals a useful result: total revenue is greatest at the quantity where marginal revenue is zero. This is a revenue result only; it does not, by itself, determine the output a firm should choose when production costs matter.

Definition: A linear inverse-demand function has the form $P(Q)=a-bQ$, where $P$ is price per unit, $Q$ is quantity, $a>0$ is the price-axis intercept, and $b>0$ is the absolute value of the slope. The choke quantity is $a/b$, the quantity at which the inverse-demand price reaches zero. For a nonnegative price, the relevant range is $0\leq Q\leq a/b$.

Assume throughout that the seller charges one price for every unit sold, that the inverse-demand function applies over the quantity range being studied, and that quantity can be treated as continuous. Total revenue is price multiplied by quantity. Marginal revenue is the rate at which total revenue changes as quantity increases; it is measured in dollars per additional unit when price is in dollars per unit and quantity is in units.

$$ TR(Q)=P(Q)Q=(a-bQ)Q=aQ-bQ^2 $$

Thus total revenue is a downward-opening quadratic, not a straight line. The price on the demand curve falls as quantity rises, so selling one more unit affects revenue in two ways: the seller receives revenue from that additional unit, but the lower price applies to the units sold as well. The resulting marginal-revenue function is also linear.

Why Marginal Revenue Has Twice the Slope

Differentiate total revenue with respect to quantity. This gives the marginal-revenue curve for a continuous quantity choice:

MR(Q)=\frac{dTR}{dQ} =\frac{d}{dQ}(aQ-bQ^2) =a-2bQ

The same result can be seen by calculating the revenue change for a small increase $\Delta Q$. The exact change is

TR(Q+\Delta Q)-TR(Q) =a\Delta Q-2bQ\Delta Q-b(\Delta Q)^2

Divide by $\Delta Q$, then let the increase in quantity approach zero. The final term, $-b\Delta Q$, approaches zero, leaving $a-2bQ$. This derivation explains why marginal revenue is not simply the price of the extra unit: the price reduction on existing sales is also reflected in the change in total revenue.

Theorem: If inverse demand is $P(Q)=a-bQ$, with $a>0$ and $b>0$, then $TR(Q)=aQ-bQ^2$ and $MR(Q)=a-2bQ$. Demand and marginal revenue have the same vertical intercept, $a$, but marginal revenue has twice the absolute slope. Their quantity-axis intercepts are $a/b$ and $a/(2b)$, respectively.

The quantity-axis intercept of demand is the choke quantity: $P(Q)=0$ at $Q=a/b$. Marginal revenue reaches zero halfway to that quantity, at $Q=a/(2b)$. Over the range from zero to the revenue-maximizing quantity, marginal revenue is positive and total revenue rises. Beyond that point, marginal revenue is negative and total revenue falls, even though the demand price remains nonnegative until the choke quantity.

The Revenue-Maximizing Quantity

The total-revenue maximum occurs where marginal revenue is zero. For this linear model, the result follows directly from the marginal-revenue formula. It is a maximum because the second derivative of total revenue is $-2b$, which is negative when $b>0$. Total revenue is zero at both ends of the nonnegative-price range, $Q=0$ and $Q=a/b$, so the interior maximum is the largest revenue on that range.

MR(Q)=0 \quad\Longrightarrow\quad a-2bQ=0 \quad\Longrightarrow\quad Q_R=\frac{a}{2b}

At $Q_R$, the price and total revenue are:

P(Q_R)=a-b\left(\frac{a}{2b}\right)=\frac{a}{2}, \qquad TR_{\max}=P(Q_R)Q_R =\frac{a}{2}\frac{a}{2b} =\frac{a^2}{4b}

The notation $Q_R$ here labels the quantity that maximizes revenue. It does not mean that a profit-maximizing firm must choose that quantity. A firm’s output decision also depends on costs; the revenue calculation alone says nothing about whether producing a particular quantity is profitable.

Worked Examples

Worked Example: Finding Revenue and Its Maximum

A hypothetical streaming service faces inverse demand $P(Q)=90-2Q$, where $P$ is dollars per subscription period and $Q$ is the number of subscription units, measured continuously for this model. Find the choke quantity, total-revenue and marginal-revenue functions, and the price and revenue at the revenue maximum.

Method. First identify the nonnegative-price range, then derive the two revenue functions and use $MR=0$. This avoids treating the demand curve itself as the marginal-revenue curve.

1. Choke quantity. Set price equal to zero: $90-2Q=0$, so $Q=45$. The relevant range is $0\leq Q\leq45$.

2. Total revenue. Multiply price by quantity:

$$ TR(Q)=(90-2Q)Q=90Q-2Q^2 $$

3. Marginal revenue. Differentiate total revenue:

$$ MR(Q)=90-4Q $$

4. Revenue-maximizing quantity. Set marginal revenue equal to zero: $90-4Q=0$, giving $Q_R=22.5$. This is within the feasible range.

5. Price and revenue at that quantity. Price is $90-2(22.5)=45$ dollars. Total revenue is $45(22.5)=1{,}012.50$ dollars per subscription period.

Check. Substitution into the total-revenue function gives $90(22.5)-2(22.5)^2=2{,}025-1{,}012.5=1{,}012.5$. Marginal revenue is $90-4(22.5)=0$, as required.

Worked Example: Two Quantities With the Same Revenue

A hypothetical equipment rental market has inverse demand $P(Q)=100-4Q$. Find the quantities that produce total revenue of $\$576$, check that they are feasible, and identify the revenue-maximizing quantity.

Method. Set the quadratic total-revenue function equal to the target and factor the resulting equation. Checking each root against the demand range prevents accepting an algebraic solution that implies a negative price.

1. Find the feasible range and revenue function. The choke quantity is $100/4=25$, so $0\leq Q\leq25$. Total revenue is

$$ TR(Q)=(100-4Q)Q=100Q-4Q^2 $$

2. Solve for the target revenue. Set $TR(Q)=576$:

100Q-4Q^2=576 \quad\Longrightarrow\quad Q^2-25Q+144=0 \quad\Longrightarrow\quad (Q-9)(Q-16)=0

The two quantities are $Q=9$ and $Q=16$. Both lie between zero and the choke quantity of 25. At $Q=9$, price is $100-4(9)=64$, and revenue is $64(9)=576$. At $Q=16$, price is $100-4(16)=36$, and revenue is $36(16)=576$.

3. Locate the maximum. Here $MR(Q)=100-8Q$. Setting it equal to zero gives $Q_R=12.5$, with price $100-4(12.5)=50$ and revenue $50(12.5)=625$. The target-revenue quantities are equally spaced around 12.5, and the maximum revenue exceeds the target.

Check. The revenue function gives $100(12.5)-4(12.5)^2=1{,}250-625=625$. Both target roots were checked directly in the original price-times-quantity expression.

Worked Example: Recovering a Linear Demand Curve From Two Observations

In a hypothetical market for equipment-maintenance hours, the observed price is $\$72$ per hour at quantity $Q=6$, and $\$40$ per hour at $Q=14$. Assume inverse demand is linear. Find its equation, the marginal-revenue function, and the quantity, price, and revenue at the revenue maximum.

Method. Use the two observations to find the slope and intercept of inverse demand, then apply the linear-demand revenue results. A common mistake is to use the change in price per unit of quantity as a positive demand slope.

1. Find the slope. The change in price is $40-72=-32$, while the change in quantity is $14-6=8$. Therefore the slope is $-32/8=-4$, so demand has the form $P(Q)=a-4Q$.

2. Find the intercept. Substitute the observation $(Q,P)=(6,72)$: $72=a-4(6)=a-24$, so $a=96$. Thus inverse demand is $P(Q)=96-4Q$. Checking the second observation gives $96-4(14)=40$, as stated.

3. Derive revenue and marginal revenue.

TR(Q)=(96-4Q)Q=96Q-4Q^2, \qquad MR(Q)=96-8Q

4. Find the revenue maximum. Setting $MR(Q)=0$ gives $96-8Q=0$, so $Q_R=12$. The choke quantity is $96/4=24$, confirming that 12 is feasible. At $Q=12$, price is $96-4(12)=48$ dollars per hour, and revenue is $48(12)=576$ dollars.

Check. The revenue function gives $96(12)-4(12)^2=1{,}152-576=576$, and marginal revenue is $96-8(12)=0$. The demand equation also reproduces both observations.

Reading the Curves and Avoiding Traps

The geometry offers quick checks on algebra. Demand and marginal revenue share the price-axis intercept $a$, while marginal revenue reaches the quantity axis at half the choke quantity. Total revenue rises as long as marginal revenue is positive, reaches its maximum when marginal revenue is zero, and falls when marginal revenue is negative. These statements concern revenue, not profit.

TR(Q)=\frac{a^2}{4b} -b\left(Q-\frac{a}{2b}\right)^2

This completed-square form makes the symmetry of revenue around $Q_R=a/(2b)$ explicit. Two quantities equally far from $Q_R$ have the same total revenue, provided both are within the feasible range. The revenue maximum is the value at the center, where the squared term is zero.

A frequent exam trap is solving a target-revenue equation but failing to substitute the proposed quantities back into $P(Q)Q$. For $P(Q)=100-4Q$, claiming that $Q=8$ and $Q=17$ each produce $\$576$ is incorrect: at $Q=8$, price is $68$ and revenue is $68(8)=544$; at $Q=17$, price is $32$ and revenue is $32(17)=544$. Solving $100Q-4Q^2=576$ gives $(Q-9)(Q-16)=0$, so the correct quantities are 9 and 16. Both are feasible because each is below the choke quantity of 25. The maximum occurs at $Q=12.5$, not at either target-revenue quantity.

Do not double the wrong slope. If demand has slope $-b$, marginal revenue has slope $-2b$, with the same vertical intercept.
Check feasibility. A quantity above $a/b$ implies a negative demand price in this model and is outside the nonnegative-price range.
Keep revenue separate from profit. The condition $MR=0$ identifies maximum revenue, not a cost-informed profit-maximizing output.
Respect the continuous-quantity assumption. If output must be an integer, the continuous revenue maximum may fall between feasible quantities; compare the neighboring integer quantities using total revenue.

Key takeaway: With linear inverse demand $P(Q)=a-bQ$, total revenue is $aQ-bQ^2$, marginal revenue is $a-2bQ$, and maximum revenue occurs at $Q=a/(2b)$. At that quantity, price is $a/2$ and total revenue is $a^2/(4b)$.

Check Your Understanding

Work each problem from the stated demand information, and check proposed quantities against the nonnegative-price range.

A hypothetical market has inverse demand $P(Q)=72-3Q$. Find total revenue, marginal revenue, the choke quantity, and the quantity, price, and revenue at the revenue maximum. (Answer: $TR=72Q-3Q^2$, $MR=72-6Q$, choke quantity $24$, $Q_R=12$, price $\$36$, maximum revenue $\$432$.)
For inverse demand $P(Q)=120-5Q$, evaluate price, total revenue, and marginal revenue at $Q=8$, and state whether total revenue is rising or falling there. (Answer: price $\$80$, total revenue $\$640$, $MR=\$40$ per additional unit, so revenue is rising.)
For $P(Q)=84-2Q$, find both quantities that yield total revenue of $\$800$, and check their feasibility. (Answer: $Q=21-\sqrt{41}$ and $Q=21+\sqrt{41}$, approximately $14.597$ and $27.403$; both are feasible because the choke quantity is $42$.)
A linear inverse-demand curve passes through $(Q,P)=(4,68)$ and $(10,44)$. Find the demand equation and its revenue-maximizing quantity, price, and revenue. (Answer: $P(Q)=84-4Q$, $Q_R=10.5$, price $\$42$, maximum revenue $\$441$.)
For $P(Q)=90-3Q$, compare marginal revenue and total revenue at $Q=20$ with their values at the revenue maximum. (Answer: the choke quantity is $30$; at $Q=20$, price is $\$30$, $MR=-\$30$ per additional unit, and revenue is $\$600$. The maximum occurs at $Q=15$, where price is $\$45$, $MR=0$, and revenue is $\$675$; revenue is falling at $Q=20$.)

What You'll Learn