Why the Marginal-Cost Rule Maximizes Profit

Why Compare Price With Marginal Cost?

A price-taking firm treats the market price as given. When it considers a small change in output, the key question is whether the additional revenue exceeds the additional cost. If price is greater than the cost of producing the next increment, producing more raises profit. If price is lower, reducing output raises profit. The marginal-cost rule is therefore a consequence of profit maximization, not a separate rule to memorize without conditions.

The rule identifies an interior candidate for the best output, but it does not by itself prove that the candidate is a maximum. In particular, a firm needs to distinguish an intersection where marginal cost is rising from one where it is falling. If marginal cost is not monotone, there may be several intersections, and the firm must compare the relevant candidates and feasible boundaries.

Definition: A price-taking firm sells each unit at the given market price $p$, so its total revenue at output $q$ is $R(q)=pq$. Its profit is $\pi(q)=pq-C(q)$, where $C(q)$ is total cost. When $C$ is differentiable, marginal cost is $MC(q)=C'(q)$. An interior output is a feasible output strictly between the lower and upper boundaries of the firm’s choice set.

This analysis assumes the firm takes $p$ as constant when it changes its own output, and that the cost function accurately reflects the costs that vary with output. The marginal-cost rule does not describe a firm that can affect the price it receives. It also does not remove the need to check output boundaries or compare multiple stationary points.

Deriving the Marginal-Cost Condition

For a price-taking firm, revenue changes by $p$ dollars for each additional unit. The change in total cost from a small output increase is described by marginal cost. Differentiating profit gives the precise comparison.

\pi(q)=pq-C(q), \qquad \pi'(q)=p-C'(q)=p-MC(q).

At an interior optimum where profit is differentiable, the first-order necessary condition is $\pi'(q^*)=0$. Thus $p=MC(q^*)$. This equality says that, at the candidate output, a small increase or decrease has no first-order effect on profit. It does not by itself establish that profit is maximized.

Write profit.
For a price-taking firm, $\pi(q)=pq-C(q)$, with price $p$ held constant.

Differentiate with respect to output.
The derivative is $\pi'(q)=p-MC(q)$.

Interpret the sign.
If $p>MC(q)$, then $\pi'(q)>0$ and a small increase in output raises profit. If $p<MC(q)$, then $\pi'(q)<0$ and a small decrease in output raises profit.

Check whether the candidate is a maximum.
At an interior stationary point, examine how marginal cost behaves nearby, and compare other candidates and boundaries when needed.

The sign argument also has a direct finite-change version. If cost is differentiable between two outputs $q_1$ and $q_2$, with $q_2>q_1$, then the change in profit is the integral of marginal revenue minus marginal cost over that output interval.

\pi(q_2)-\pi(q_1) =\int_{q_1}^{q_2}\bigl[p-MC(q)\bigr]\,dq.

This equality follows from the fundamental theorem of calculus because $\pi'(q)=p-MC(q)$. If marginal cost is below price throughout the interval, the integrand is positive and the increase in output raises profit. If marginal cost is above price throughout, the integral is negative and the increase lowers profit. This explains why a firm should expand when price exceeds marginal cost and contract when price is below marginal cost.

Why Rising Marginal Cost Matters

Suppose $C$ is twice differentiable. At an interior stationary point $q^*$, the second derivative of profit is $\pi''(q^*)=-C''(q^*)$. For a local maximum, the second-derivative necessary condition is $\pi''(q^*)\leq 0$, which implies $C''(q^*)\geq 0$. Since $C''$ is the derivative of marginal cost, this condition means marginal cost cannot be falling at that point. It does not require strictly rising marginal cost.

If $C''(q^*)>0$, then $\pi''(q^*)<0$, which is a sufficient second-order condition for a strict local maximum. If $C''(q^*)=0$, the second-derivative test is inconclusive: the point may still be a strict maximum, or it may not be. A rising marginal-cost curve near the price intersection is a useful way to see the relevant direction: below the intersection, price exceeds marginal cost and profit rises with output; above it, marginal cost exceeds price and profit falls.

Theorem: If marginal cost is continuous and strictly increasing over the relevant feasible output range, and it crosses the given price at an interior output $q^*$, then profit rises up to $q^*$ and falls after it. Thus $q^*$ is the unique profit-maximizing output on that range. If marginal cost is not increasing throughout the range, the equation $p=MC(q)$ can have multiple solutions; the first-order condition alone does not select the maximum.

The uniqueness result follows from the sign of $p-MC(q)$. Strictly increasing marginal cost that equals $p$ at $q^*$ is below $p$ for outputs just below $q^*$, and above $p$ for outputs just above it. If it is strictly increasing across the whole range, those signs hold throughout the respective intervals. The derivative of profit is therefore positive before $q^*$ and negative after it, so profit increases and then decreases. If a feasible boundary is included, it should still be checked unless the stated sign pattern covers the entire choice set.

Worked Examples

Worked Example: A Unique Interior Optimum

A price-taking firm can choose any output $q\geq 0$. Its market price is $p=\$20$ per unit and its cost function is $C(q)=100+4q+q^2$, in dollars. Find the profit-maximizing output and verify that it is the global maximum. Use the first- and second-derivative method.

Step 1: Find marginal cost. Differentiating the cost function gives $MC(q)=C'(q)=4+2q$. Profit is $\pi(q)=20q-(100+4q+q^2)=-100+16q-q^2$.

Step 2: Find the stationary point. The first derivative is $\pi'(q)=16-2q$. Set it equal to zero: $16-2q=0$, so $2q=16$ and $q^*=8$. Equivalently, $MC(8)=4+2(8)=20=p$.

Step 3: Check the maximum. The second derivative is $\pi''(q)=-2<0$ for every output. Therefore profit is strictly concave, and the stationary point is the unique global maximum on $q\geq0$. It is feasible because $8\geq0$. As a check, at this output revenue is $20(8)=\$160$, cost is $100+4(8)+8^2=100+32+64=\$196$, and $\pi(8)=160-196=-\$36$. The calculation identifies the output that maximizes profit; it does not require profit to be positive.

Method and mistake avoided. The method is to solve $p=MC(q)$ and then use concavity to verify the maximum. Stopping after the equality without checking curvature would leave the conclusion unjustified.

Worked Example: Two Price Intersections

A firm chooses any $q\geq0$, faces the constant price $p=\$8$, and has cost $C(q)=q^3/3-4q^2+15q+30$. Find the stationary outputs and identify the global profit maximum. This example tests why an intersection on a falling marginal-cost segment is not the maximizing one.

Step 1: Calculate marginal cost and stationary outputs. $MC(q)=C'(q)=q^2-8q+15$. The first-order condition is $8=q^2-8q+15$, or $q^2-8q+7=0$. Factoring gives $(q-1)(q-7)=0$, so the stationary outputs are $q=1$ and $q=7$.

Step 2: Apply the second-derivative test. Here $C''(q)=2q-8$, so $\pi''(q)=-C''(q)=8-2q$. At $q=1$, $\pi''(1)=8-2=6>0$, making $q=1$ a local minimum. At $q=7$, $\pi''(7)=8-14=-6<0$, making $q=7$ a strict local maximum.

Step 3: Verify the global result over feasible outputs. The profit derivative factors as $\pi'(q)=8-MC(q)=-q^2+8q-7=-(q-1)(q-7)$. It is negative for $0\leq q<1$, positive for $1<q<7$, and negative for $q>7$. Thus profit falls from the boundary $q=0$ to $q=1$, rises from $q=1$ to $q=7$, and falls after $q=7$. To check that the later maximum also exceeds the boundary value, compute $\pi(7)-\pi(0)=\int_0^7(-q^2+8q-7)\,dq$ $=[-q^3/3+4q^2-7q]_0^7=-343/3+196-49=98/3>0$. Therefore $q=7$ is the global maximum on $q\geq0$.

Method and mistake avoided. The method is to find every solution to $p=MC(q)$, classify each using curvature, and check the boundary when establishing a global maximum. Treating every price–marginal-cost intersection as a maximum would incorrectly select $q=1$.

Worked Example: Whole-Unit Output With a Restricted Choice Set

A hypothetical producer sells output at $p=\$18$ per unit and must choose a whole-number output from 0 through 5 units. The marginal costs of units 1 through 5, respectively, are $\$7,\$11,\$15,\$19,\$24$. Find the best output among the stated choices. Assume the listed marginal costs are the cost of each successive unit.

Step 1: Compare the extra revenue and cost of each unit. Each extra unit adds $\$18$ of revenue. The first three units have marginal costs below $\$18$, so each raises profit by, respectively, $18-7=\$11$, $18-11=\$7$, and $18-15=\$3$. The fourth unit changes profit by $18-19=-\$1$, and the fifth by $18-24=-\$6$.

Step 2: Accumulate the changes for all feasible outputs. Total cost for outputs 0 through 5 is respectively $\$0,\$7,\$18,\$33,\$52,\$76$, found by cumulatively adding the given unit costs. Revenue at those outputs is respectively $\$0,\$18,\$36,\$54,\$72,\$90$. Thus revenue less the listed output-dependent cost is $\$0,\$11,\$18,\$21,\$20,\$14$, respectively. The greatest value is $\$21$, at 3 units.

Conclusion and check. The best output among the choices from 0 through 5 is 3 units. The fourth and fifth units reduce the objective relative to 3, as their individual costs exceed their additional revenue. Restricting the choice set matters: the data establish the best choice among 0 through 5, not among unspecified outputs beyond 5.

Method and mistake avoided. For whole-unit decisions, compare price with the marginal cost of each added unit, then check the cumulative values over the stated feasible set. Claiming an optimum over outputs beyond the stated range would require additional cost information.

Common Exam Traps

Assuming $p=MC$ proves a maximum. It is an interior first-order condition. With falling marginal cost, the stationary point may instead be a minimum.
Requiring strictly positive curvature as a necessary condition. At a twice-differentiable local maximum, $C''(q^*)\geq0$ is necessary; $C''(q^*)>0$ is sufficient for a strict local maximum. Equality makes the second-derivative test inconclusive.
Applying the continuous condition to a whole-unit choice without adjustment. In a discrete problem, compare the extra revenue and marginal cost for each feasible unit and evaluate the stated output choices.
Ignoring the range of feasible output. A schedule for a finite number of units can establish an optimum only over that specified range unless information about later units is also supplied.
Forgetting that the firm takes price as given. The derivative of revenue is $p$ here because price is constant with respect to the firm’s output choice.

Key takeaway: For a price-taking firm, $\pi'(q)=p-MC(q)$: price above marginal cost gives an incentive to expand, and price below marginal cost gives an incentive to contract. At an interior optimum, $p=MC$; a rising marginal-cost condition supports a maximum, but multiple stationary points and feasible boundaries must be checked when the cost curve does not give a unique increasing crossing.

Check Your Understanding

For each problem, identify the relevant marginal-cost comparison and check whether the proposed output is a maximum over the stated choice set.

A price-taking firm chooses any $q\geq0$, faces $p=\$26$, and has $C(q)=50+6q+q^2$. Find its maximizing output and state why it is a maximum. (Answer: $q=10$; $MC=6+2q$, and profit is strictly concave.)
A firm chooses $q\geq0$, faces $p=\$8$, and has $C(q)=q^3/3-4q^2+15q+30$. Which stationary output is a local minimum and which is a local maximum? (Answer: $q=1$ is a local minimum; $q=7$ is a strict local maximum.)
A hypothetical price-taking producer chooses a whole-number output from 0 through 4 units. Price is $\$15$, and the marginal costs of units 1 through 4 are $\$5,\$10,\$15,\$19$. Compare the cumulative output-dependent profit contributions and give the best output. (Answer: 3 units; the contributions for outputs 0 through 4 are $\$0,\$10,\$15,\$15,\$11$, so outputs 2 and 3 tie.)
A firm has $C(q)=pq+(q-2)^4+K$, where $p$ and $K$ are constants, and can choose any $q\geq0$. Determine its maximizing output and explain why a zero second derivative there does not rule out a strict maximum. (Answer: $q=2$ is the strict global maximum because $\pi(q)=-(q-2)^4-K\leq-K=\pi(2)$, with equality only at $q=2$; $C''(2)=0$, so the second-derivative test is inconclusive.)

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