Why Marginal Revenue Must Equal Marginal Cost
A firm deciding how much to produce compares the extra revenue from a small increase in output with the extra cost of producing it. If the extra revenue exceeds the extra cost, increasing output raises profit; if it is smaller, reducing output raises profit. At an interior optimum, where output can be adjusted slightly in either direction, neither change should improve profit. This reasoning leads to the condition marginal revenue equals marginal cost.
The condition is a first-order requirement, not a complete test that an output actually maximizes profit. It applies when revenue and cost are differentiable and the candidate output is interior. In particular, satisfying the equation alone does not establish that profit is at its highest possible level.
Deriving the First-Order Condition
Suppose a firm chooses a positive output \(q^*\), and \(q^*\) is an interior local profit maximum. “Interior” means output can be adjusted by a small amount both upward and downward while remaining feasible. Assume \(R\) and \(C\) are differentiable at \(q^*\). Profit is the difference between revenue and cost, so its derivative follows by differentiating each term.
Since the derivative of a difference is the difference of the derivatives, \(\pi'(q)=R'(q)-C'(q)\).
At a differentiable interior local maximum, the first derivative of profit must be zero: \(\pi'(q^*)=0\).
Because \(R'(q)=MR(q)\) and \(C'(q)=MC(q)\), the zero-derivative equation becomes \(MR(q^*)-MC(q^*)=0\).
The reason the derivative must be zero is that a nonzero slope indicates a locally profitable direction of adjustment. If \(\pi'(q^*)>0\), a sufficiently small increase in output raises profit. If \(\pi'(q^*)<0\), a sufficiently small decrease raises profit. Neither is consistent with an interior local maximum. This argument requires differentiability and two-sided feasible adjustments; it does not assert that every solution of \(MR=MC\) is a maximum.
The marginal quantities are measured in dollars per unit of output, so they can be compared directly. Total revenue and total cost, by contrast, are dollar amounts. Equating total revenue with total cost is a break-even condition, not the marginal condition derived here.
Using Revenue and Cost Functions
When inverse demand gives the price \(P(q)\) received for each unit, total revenue is \(R(q)=P(q)q\). It must be differentiated as a product. Thus, when \(P\) is differentiable, marginal revenue is \(R'(q)=P(q)+qP'(q)\), not generally just the price \(P(q)\). Total cost is differentiated in the same way to obtain marginal cost.
A fixed cost appears in total cost but disappears when taking the derivative with respect to output: the derivative of a constant is zero. It can still affect the level of profit at a chosen output. This distinction is important: a fixed cost does not change the interior equation \(MR=MC\), but it does change calculated profit.
Worked Example: Solving a Quadratic Profit Condition
A hypothetical producer has revenue \(R(q)=110q-3q^2\) dollars and total cost \(C(q)=90+20q+q^2\) dollars, where \(q\) is output in units. Find the output satisfying the interior first-order condition, then calculate profit there and check the marginal values.
Method. Differentiate total revenue and total cost, equate their derivatives, and then substitute the resulting output into both functions. The common mistake avoided here is omitting fixed cost when calculating profit, even though it has no effect on marginal cost.
Step 1: Find marginal revenue. Differentiating \(R(q)=110q-3q^2\) gives \(MR(q)=110-6q\).
Step 2: Find marginal cost. Differentiating \(C(q)=90+20q+q^2\) gives \(MC(q)=20+2q\). The fixed cost of 90 has derivative zero.
Step 3: Solve the condition. Set \(110-6q=20+2q\). Then \(90=8q\), so \(q=11.25\) units.
Step 4: Calculate revenue. At \(q=11.25\), \(q^2=126.5625\). Therefore \(R(11.25)=110(11.25)-3(126.5625)=1237.5-379.6875=\$857.8125\).
Step 5: Calculate cost and profit. Total cost is \(C(11.25)=90+20(11.25)+126.5625=90+225+126.5625=\$441.5625\). Profit is \(857.8125-441.5625=\$416.25\). Reported to the nearest cent, profit is \(\$416.25\).
Check. \(MR(11.25)=110-67.5=\$42.50\) per unit, and \(MC(11.25)=20+22.5=\$42.50\) per unit. The marginal values agree, as required by the first-order condition.
Worked Example: Including a Fixed Cost in Profit
A hypothetical firm has \(R(q)=96q-2q^2\) dollars and \(C(q)=150+24q+0.5q^2\) dollars. Find the output that satisfies \(MR=MC\), calculate revenue, cost, and profit at that output, and verify the marginal equality.
Method. Differentiate the functions first, solve the linear equation, and then evaluate the original total revenue and cost functions. The common mistake avoided is calculating profit from marginal values or dropping the fixed cost when evaluating total cost.
Step 1: Differentiate. Marginal revenue is \(MR(q)=96-4q\). Marginal cost is \(MC(q)=24+q\), because the fixed cost 150 contributes zero to the derivative.
Step 2: Solve for output. Set \(96-4q=24+q\). Then \(72=5q\), so \(q=14.4\) units.
Step 3: Evaluate revenue. Since \(14.4^2=207.36\), \(R(14.4)=96(14.4)-2(207.36)=1382.4-414.72=\$967.68\).
Step 4: Evaluate cost. \(C(14.4)=150+24(14.4)+0.5(207.36)=150+345.6+103.68=\$599.28\).
Step 5: Calculate profit and check. Profit is \(967.68-599.28=\$368.40\). Also, \(MR(14.4)=96-57.6=\$38.40\) per unit and \(MC(14.4)=24+14.4=\$38.40\) per unit. The profit arithmetic and marginal condition both check.
Worked Example: Deriving Marginal Revenue from Inverse Demand
A hypothetical seller faces inverse demand \(P(q)=84-1.5q\) dollars per unit and has total cost \(C(q)=240+12q+0.75q^2\) dollars. Find the output satisfying the interior first-order condition, then compute the price and profit at that output.
Method. First construct total revenue from price times quantity, then differentiate revenue and cost. The common mistake avoided is setting price equal to marginal cost without first deriving marginal revenue from the demand function.
Step 1: Construct total revenue. \(R(q)=P(q)q=(84-1.5q)q=84q-1.5q^2\).
Step 2: Differentiate revenue and cost. \(MR(q)=84-3q\). Also, \(MC(q)=12+1.5q\), since the fixed cost 240 has zero derivative.
Step 3: Solve the first-order condition. Set \(84-3q=12+1.5q\). Thus \(72=4.5q\), giving \(q=16\) units.
Step 4: Find price, revenue, and cost. Price is \(P(16)=84-1.5(16)=84-24=\$60\) per unit, so \(R(16)=60(16)=\$960\). Cost is \(C(16)=240+12(16)+0.75(16^2)=240+192+192=\$624\).
Step 5: Calculate profit and check. Profit is \(960-624=\$336\). At this output, \(MR(16)=84-48=\$36\) per unit and \(MC(16)=12+24=\$36\) per unit. The price is \(\$60\), not \(\$36\); it is marginal revenue, rather than price, that equals marginal cost in this calculation.
What the Condition Does—and Does Not—Tell You
The equation \(MR=MC\) identifies an output at which the first derivative of profit is zero, provided the differentiability and interior conditions hold. It is therefore a way to find a candidate output. It is not, by itself, proof that the candidate maximizes profit: the first-order equation can have solutions that are not profit maxima, and a maximum may not be interior. Establishing whether a candidate is a maximum requires additional analysis.
The condition is useful because it expresses the output decision in marginal terms. At an output where \(MR>MC\), the derivative of profit is positive, so a small increase in output raises profit. Where \(MR<MC\), the derivative is negative, so a small decrease raises profit. At an interior stationary point, the marginal gain and marginal cost of a small output change balance at the margin.
| Expression | Meaning | Role in the decision |
|---|---|---|
| \(R(q)=C(q)\) | Total revenue equals total cost | Break-even; not the marginal first-order condition |
| \(MR(q)=MC(q)\) | The derivatives of revenue and cost are equal | Necessary first-order condition for an interior local maximum |
| \(P(q)=MC(q)\) | Price equals marginal cost | Not generally valid; it requires a revenue setting in which \(MR=P\) |
Common Mistakes to Avoid
- Equating totals. Do not set \(R(q)=C(q)\) to find the output satisfying the marginal condition. Differentiate each total function and set \(R'(q)=C'(q)\).
- Using price as marginal revenue automatically. With output-dependent price, use \(R(q)=P(q)q\) and differentiate. The derivative generally differs from price.
- Keeping fixed cost in marginal cost. A constant fixed cost affects total cost and profit, but its derivative with respect to output is zero.
- Stopping after solving the equation. The equality gives a first-order candidate, not a full demonstration of maximum profit.
- Mixing units. \(R\) and \(C\) are total dollar amounts; \(MR\) and \(MC\) are dollars per unit. Compare marginal values with each other, not with total values.
Check Your Understanding
For each problem, use the interior first-order condition and check any requested total values.
- A hypothetical workshop has \(R(q)=72q-1.5q^2\) and \(C(q)=80+12q+0.5q^2\), in dollars. Find the output satisfying \(MR=MC\), calculate profit there, and verify the marginal values. (Answer: \(q=15\) units; profit \(=\$370.00\); \(MR=MC=\$27\) per unit.)
- A seller faces inverse demand \(P(q)=100-2q\) and has cost \(C(q)=100+20q+q^2\), in dollars. Find the output satisfying the condition and calculate profit, giving profit to the nearest cent. (Answer: \(q=40/3\) units; profit \(=\$433.33\).)
- A hypothetical service has revenue \(R(q)=90q-2.5q^2\) and cost \(C(q)=60+15q+0.5q^2\), in dollars. Find the first-order candidate, evaluate revenue and cost there, and check marginal equality. (Answer: \(q=12.5\) units; revenue \(=\$734.375\); cost \(=\$325.625\); profit \(=\$408.75\); \(MR=MC=\$27.50\) per unit.)
- At an output of 8 units, a firm has \(MR=\$31\) per unit and \(MC=\$27\) per unit. Using the derivative of profit, state the sign of \(\pi'(8)\) and whether a small increase in output raises or lowers profit. (Answer: \(\pi'(8)=\$4\) per unit; a small increase raises profit.)
- A firm has \(R(q)=60q-q^2\) and \(C(q)=100+10q+q^2\), in dollars. Solve the first-order equation and calculate profit at its solution. State whether the equation alone establishes that the solution is a maximum. (Answer: \(q=12.5\) units; profit \(=\$212.50\); no, the first-order equation alone does not establish a maximum.)