What Is Marginal Cost And Marginal Revenue?
Marginal cost and marginal revenue are the two derivatives a firm uses to decide how much to produce. If $C(x)$ is the total cost of making $x$ units and $R(x)$ is the total revenue from selling them, then the marginal functions are their derivatives:
$$MC = C'(x), \qquad MR = R'(x)$$
Marginal cost is the rate at which total cost rises as output grows. Marginal revenue is the rate at which total revenue rises as sales grow. Because a derivative is a rate of change, each one answers a per-unit question at the current level of production, not an average taken over all units so far.
Profit is what the firm actually cares about, and it is the gap between the two totals:
$$P(x) = R(x) - C(x)$$
Differentiating gives the marginal profit, the extra profit from one more unit:
$$P'(x) = R'(x) - C'(x) = MR - MC$$
That single line drives the whole subject. When $MR > MC$ the next unit adds profit, when $MR < MC$ it subtracts profit, and the best output sits where the two are equal. This is one of the cleanest places calculus meets money, and it sits inside the wider field of calculus in economics.
Throughout this article the derivative is written in prime notation: $C'(x)$, $R'(x)$, and $P'(x)$.
Why Is Marginal Cost "The Cost Of One More Unit"?
The phrase every textbook uses is that marginal cost is the cost of producing one additional unit. That is a plain-English reading of a derivative, and it is worth seeing exactly why the two agree.
The derivative is a limit of average changes:
$$C'(x) = \lim_{h \to 0} \frac{C(x+h) - C(x)}{h}$$
The smallest meaningful step in production is one whole unit, so set $h = 1$. The limit has not been taken yet, but for a smooth cost curve the value at $h = 1$ is very close to the true slope:
$$C'(x) \approx \frac{C(x+1) - C(x)}{1} = C(x+1) - C(x)$$
The right-hand side is the exact cost of the $(x+1)$th unit. So $C'(x)$, the marginal cost, approximates the cost of the next unit, and the two are usually so close that firms treat them as the same number. The same reading gives marginal revenue as the money the next sale brings in.
Geometrically, $C'(x)$ is the slope of the tangent line to the total-cost curve at $x$. A steep cost curve means each new unit is expensive; a gently rising curve means the next unit is cheap. Pairing the algebra ($C'(x)$) with the geometry (tangent slope) is the same move used all through applications of derivatives.
How Do You Compute Marginal Cost, Revenue, And Profit? Worked Examples
Each example is fully stepped, and every optimum is checked with a second derivative.
Example 1: Marginal cost as the cost of the next unit.
A workshop has total cost $C(x) = 0.01x^2 + 5x + 500$ dollars for $x$ items. Find the marginal cost at $x = 100$ and compare it with the true cost of the 101st item.
Differentiate with the power rule:
$$C'(x) = 0.02x + 5, \qquad C'(100) = 0.02(100) + 5 = 7$$
Now the exact cost of the 101st item:
$$C(101) - C(100) = 1107.01 - 1100 = 7.01$$
The marginal cost $$7$ predicts the true next-unit cost $$7.01$ to within a cent.
Final answer: $C'(100) = $7$ per unit, matching the actual next-unit cost of $$7.01$.
Example 2: Marginal revenue from a demand function.
When price depends on quantity, revenue is price times quantity, so it needs the product rule or a quick expansion. Suppose the demand (price) function is $p = 20 - 0.1x$. Then
$$R(x) = p \cdot x = (20 - 0.1x)x = 20x - 0.1x^2$$
$$R'(x) = 20 - 0.2x, \qquad R'(50) = 20 - 0.2(50) = 10$$
At $x = 50$ the marginal revenue is $$10$, yet the price there is $p = 20 - 0.1(50) = $15$. The next sale brings in less than the sticker price, because selling one more unit forces the price down on every unit. Confusing these two is a classic slip, covered in the mistakes section.
Final answer: $R'(50) = $10$ per unit, below the price of $$15$.
Example 3: Profit maximization with $MR = MC$.
A firm has $C(x) = 0.1x^2 + 6x + 200$ and faces demand $p = 30 - 0.2x$. Find the output that maximizes profit.
First the marginal functions:
$$MC = C'(x) = 0.2x + 6, \qquad R(x) = 30x - 0.2x^2, \qquad MR = R'(x) = 30 - 0.4x$$
Set $MR = MC$ and solve:
$$30 - 0.4x = 0.2x + 6 ;\Longrightarrow; 24 = 0.6x ;\Longrightarrow; x = 40$$
Confirm it is a maximum with the profit function $P(x) = R(x) - C(x) = 24x - 0.3x^2 - 200$:
$$P'(x) = 24 - 0.6x, \qquad P''(x) = -0.6 < 0$$
The negative second derivative means the curve bends downward, so $x = 40$ is a maximum. The profit there is $P(40) = 24(40) - 0.3(40)^2 - 200 = 960 - 480 - 200 = 280$.
Final answer: produce $x = 40$ units for a maximum profit of $$280$ (selling price $p = $22$).
Why Does $MR = MC$ Maximize Profit?
The rule falls straight out of the marginal profit $P'(x) = MR - MC$, and the geometry makes it obvious.
Below the optimum, $MR > MC$. Marginal profit is positive, the profit curve is still climbing, and each extra unit adds money. The firm should keep producing.
Above the optimum, $MR < MC$. Marginal profit is negative, the profit curve is falling, and each extra unit destroys money. The firm has gone too far.
At the optimum, $MR = MC$. Marginal profit is zero, the tangent to the profit curve is flat, and profit is neither rising nor falling. This is the turning point.
A flat tangent is exactly the condition for a maximum or minimum, which is why $MR = MC$ is a statement about the derivative as a rate of change equalling zero. The subtlety is that a flat tangent can also sit at the bottom of a dip, so the equation alone is not the full story.
When Does $MR = MC$ Fail To Give A Maximum?
Setting $MR = MC$ finds every flat spot on the profit curve, and some flat spots are minima. This is where the second-order check earns its keep, and where many students lose marks.
Take constant price so $R(x) = 200x$ and $MR = 200$, with a cubic cost $C(x) = x^3 - 24x^2 + 200x + 100$. Then
$$MC = C'(x) = 3x^2 - 48x + 200$$
Setting $MR = MC$ gives $3x^2 - 48x + 200 = 200$, so $3x(x - 16) = 0$ and $x = 0$ or $x = 16$. Two outputs satisfy the rule. The profit function decides between them:
$$P(x) = 200x - C(x) = -x^3 + 24x^2 - 100, \qquad P''(x) = -6x + 48$$
At $x = 0$, $P''(0) = 48 > 0$, a profit minimum. At $x = 16$, $P''(16) = -48 < 0$, a profit maximum, worth $P(16) = 1948$. The genuine profit peak is $x = 16$; the other root is the worst nearby choice.
The reliable rule is therefore two-part: solve $MR = MC$, then keep only the root where $P''(x) < 0$ (equivalently, where marginal cost cuts up through marginal revenue). This is the second derivative test applied to money.
What Are The Total And Marginal Functions At A Glance?
Every quantity in the topic is a total function paired with its derivative. Keeping the pairs straight prevents most errors.
Table: Total functions and their marginal counterparts in calculus.
Quantity | Total function | Marginal function (derivative) | What it measures |
|---|---|---|---|
Cost | $C(x)$ | $MC = C'(x)$ | cost of the next unit |
Revenue | $R(x)$ | $MR = R'(x)$ | revenue from the next sale |
Profit | $P(x) = R(x) - C(x)$ | $MP = P'(x) = MR - MC$ | profit from the next unit |
Average cost | $\dfrac{C(x)}{x}$ | (not a marginal) | cost spread over all units so far |
The last row is the trap. Average cost divides the total by every unit made; marginal cost differentiates the total to isolate the next unit. They answer different questions and rarely give the same number.
Why Does Marginal Analysis Work? The Intuition
Marginal thinking works because a total is built one unit at a time, and calculus is the tool for "one more."
A total is an accumulation of marginals. Total cost is the running sum of the cost of each successive unit, so the rate at which that sum grows is precisely the marginal cost. The rate of change of the total is the marginal value.
Decisions live at the edge, not the average. Whether to make the next unit depends only on what that unit adds and costs, not on units already produced. Sunk totals do not change the marginal comparison.
The optimum is a balance point. Profit stops growing exactly when the money from the next unit ($MR$) matches its cost ($MC$). Past that point the firm pays more than it earns for each additional unit.
Seen this way, marginal cost and marginal revenue are not extra formulas layered on economics. They are the derivative doing what it always does, measuring the effect of a small change, applied to cost and revenue.
Who Discovered Marginal Cost And Marginal Revenue Analysis?
Marginal analysis is where nineteenth-century calculus quietly took over economics, decades before most economists were comfortable with derivatives.
Two later waves built on the idea:
The Marginal Revolution (1870s). William Stanley Jevons (1835–1882, England), Carl Menger (1840–1921, Austria), and Léon Walras (1834–1910, France) independently rebuilt economic value around marginal quantities, making the derivative central to the subject.
Alfred Marshall (1842–1924, England) turned the marginal-cost and marginal-revenue curves into the supply-and-demand diagrams still taught today, cementing $MR = MC$ as the standard rule for output.
Where Is Marginal Cost And Marginal Revenue Used In The Real World?
The MR-versus-MC comparison runs quietly through decisions far beyond a textbook firm.
Manufacturing and operations: production planners set output near the point where the marginal cost of the next unit meets the marginal revenue it earns, avoiding both underproduction and loss-making overproduction.
Airlines and dynamic pricing: the marginal cost of one more passenger on a scheduled flight is tiny, so revenue-management systems keep selling seats while marginal revenue stays above that low marginal cost.
Electricity grids: power is dispatched in "merit order," bringing generators online from cheapest to most expensive, so the price often settles at the marginal cost of the last plant needed to meet demand.
Telecom and software: once a network or app is built, the marginal cost of an extra user is close to zero, which is why marginal-revenue thinking dominates pricing for digital goods.
Agriculture and resources: a farmer adds fertilizer or labour only while the marginal revenue of the extra yield beats the marginal cost of the input.
One derivative-based comparison, "does the next unit earn more than it costs," reappears across industries that otherwise share almost nothing.
What Are The Most Common Mistakes With Marginal Cost And Marginal Revenue?
These four errors account for most lost marks, and each matches a question real students ask on course notes, Q&A threads, and a published study of how learners read the marginal derivative.
Confusing marginal with average.
Where it slips in:
A student computes total cost divided by quantity, $C(x)/x$, and calls it the marginal cost, or divides total revenue by units sold and calls it marginal revenue.
Don't do this:
Do not average the total. Averaging spreads cost over all units; it does not isolate the next one.
The correct way:
Differentiate. Marginal cost is $C'(x)$ and marginal revenue is $R'(x)$, the rates at the current output, which differ from the averages $C(x)/x$ and $R(x)/x$.
Treating marginal revenue as the price.
Where it slips in:
On a downward-sloping demand curve, a student uses the price $p$ as the marginal revenue and skips differentiating $R(x) = p(x)\cdot x$.
Don't do this:
Do not equate marginal revenue with price when price depends on quantity. Selling one more unit lowers the price on all units.
The correct way:
Build $R(x) = p(x)\cdot x$ first, then differentiate. For $p = 20 - 0.1x$ the revenue is $20x - 0.1x^2$ and $R'(x) = 20 - 0.2x$, which is below the price at every positive $x$.
Solving $MR = MC$ and stopping there.
Where it slips in:
A student sets $MR = MC$, solves for $x$, and reports it as the profit-maximizing output without checking whether it is a peak or a dip.
Don't do this:
Do not assume every solution of $MR = MC$ maximizes profit. Some are profit minima.
The correct way:
Apply the second-order check: keep the root where $P''(x) < 0$, meaning marginal cost rises through marginal revenue. When two roots appear, this test alone tells them apart.
Reading the "one more unit" approximation as exact.
Where it slips in:
A student insists $C'(x)$ equals $C(x+1) - C(x)$ to the last cent and is thrown when the two differ slightly.
Don't do this:
Do not treat the derivative and the exact next-unit cost as identical. The derivative is the limiting slope, not the finite difference.
The correct way:
Use $C'(x) \approx C(x+1) - C(x)$ as a close estimate. In Example 1, the marginal cost $7$ approximates the true next-unit cost $7.01$, near but not identical.
Practice Problems On Marginal Cost And Marginal Revenue
Work each one, then check against the verified answer.
For $C(x) = 0.02x^2 + 8x + 400$, find the marginal cost function and $MC$ at $x = 50$.
(Answer: $C'(x) = 0.04x + 8$; $C'(50) = 10$.)Demand is $p = 50 - 0.5x$. Find $R(x)$ and the marginal revenue.
(Answer: $R(x) = 50x - 0.5x^2$; $R'(x) = 50 - x$.)A firm has $MR = 40 - 0.4x$ and $MC = 0.2x + 4$. Find the profit-maximizing output.
(Answer: $40 - 0.4x = 0.2x + 4 \Rightarrow x = 60$; marginal cost slope $0.2$ exceeds marginal revenue slope $-0.4$, so it is a maximum.)With $C(x) = x^2 + 10x + 50$ and $R(x) = 40x - x^2$, find the profit-maximizing output and the maximum profit.
(Answer: $MR = MC$ gives $40 - 2x = 2x + 10 \Rightarrow x = 7.5$; $P''(x) = -4 < 0$; maximum profit $P(7.5) = $62.50$.)Marginal profit is $P'(x) = MR - MC$. If $MR = MC$ at $x = 20$ and $P''(20) < 0$, what does $x = 20$ represent?
(Answer: the profit-maximizing output.)For $C(x) = 0.05x^2 + 3x + 120$, estimate the cost of the 21st unit using marginal cost.
(Answer: $C'(x) = 0.1x + 3$; $C'(20) = $5$, close to the exact $C(21) - C(20) = $5.05$.)
Where Should You Go Next After Marginal Cost And Marginal Revenue?
Marginal analysis opens onto the broader optimization side of calculus, and several natural doors follow.
Optimization problems. The same set-the-derivative-to-zero method applied to areas, distances, and materials, not only profit.
Applications of derivatives. The full toolkit of rates, tangents, and extrema that marginal cost and marginal revenue are one instance of.
Elasticity of demand. The next economics-and-calculus topic, measuring how sensitively quantity responds to price.
If your child is meeting these ideas for the first time, a live Bhanzu trainer teaches them from the graph up in the Bhanzu math classes.
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