Why the Marginal-Equality Rule Needs Conditions
A firm choosing output compares the additional revenue from one more unit with the additional cost of producing it. At an interior profit-maximizing output, neither increasing nor decreasing production slightly should improve profit. That balance gives the familiar condition marginal revenue equals marginal cost. But the equality alone does not establish that profit is maximized: it may identify a minimum, or a point that is not feasible.
This tutorial derives the condition precisely and explains the role of the second derivative. The distinction matters in monopoly problems because a monopolist’s marginal revenue is generally not the market price. The relevant comparison is between marginal revenue and marginal cost, evaluated at the same output.
The firm’s feasible outputs form a set, often an interval such as \(0\le q\le \bar q\). An interior choice is strictly inside that interval: small increases and small decreases in output are both feasible. The first-order condition applies to an interior optimum under differentiability. It need not hold at a boundary, where only one direction of change may be possible.
Deriving the First-Order Condition
Suppose profit is differentiable near an interior output \(q^*\), and \(q^*\) is a local maximum. For a small positive change \(h\), the output \(q^*+h\) cannot yield higher profit. For a small negative change, \(q^*+h\) cannot yield higher profit either. Dividing the profit changes by \(h\) reverses the inequality when \(h\) is negative; taking the two-sided derivative therefore requires that derivative to equal zero.
This is a necessary condition for a differentiable interior local maximum. It says that, at the chosen output, a small increase in output has zero first-order effect on profit. If \(MR>MC\), then \(\pi'>0\), so a small increase in output raises profit. If \(MR<MC\), then \(\pi'<0\), so a small decrease in output raises profit. Either strict inequality is inconsistent with an interior optimum.
How the Second-Order Condition Classifies the Point
To determine what kind of stationary point has been found, examine how the difference \(MR-MC\) changes as output moves around it. When profit is twice differentiable,
At a stationary point, a second-order Taylor expansion gives the justification. For a small change \(h\), the first-order term vanishes because \(\pi'(q^*)=0\). The leading change in profit is therefore determined by the sign of the second-order term:
Since \(h^2\) is positive for either direction, a negative second derivative makes the nearby profit change negative; a positive second derivative makes it positive. This gives the standard second-order test.
A strict local maximum is not automatically a global maximum. It is best among outputs sufficiently close to \(q^*\), but a distant feasible output could yield greater profit. If profit is strictly concave throughout the feasible interval, a stationary point is the unique global maximum on that interval. Otherwise, compare relevant stationary points and feasible boundaries.
Worked Example: A Strict Interior Profit Maximum
A monopolist faces inverse demand \(p(q)=100-q\) and has total cost \(TC(q)=20q+q^2\), for feasible output \(0\le q\le100\). Find the output and price that maximize profit, and verify that the stationary point is a maximum.
Method. Construct total revenue, equate marginal revenue and marginal cost to find an interior candidate, then check the second-order condition and profit.
Total revenue is price times quantity: \(TR(q)=q(100-q)=100q-q^2\). Therefore, \(MR(q)=TR'(q)=100-2q\). Marginal cost is \(MC(q)=TC'(q)=20+2q\).
Set marginal revenue equal to marginal cost: \(100-2q=20+2q\). Subtracting 20 and adding \(2q\) gives \(80=4q\), so \(q^*=20\). This is interior because \(0<20<100\).
The market price at this output is \(p(20)=100-20=80\). Total revenue is \(TR(20)=100(20)-20^2=2{,}000-400=1{,}600\). Total cost is \(TC(20)=20(20)+20^2=400+400=800\). Thus profit is \(\pi(20)=1{,}600-800=800\).
To classify the point, \(\pi''(q)=MR'(q)-MC'(q)=-2-2=-4<0\), so it is a strict local maximum. In fact, \(\pi(q)=80q-2q^2\) is strictly concave over the feasible interval, so this stationary point is also the unique global maximum. Check: \(MR(20)=100-40=60\) and \(MC(20)=20+40=60\), as required.
Worked Example: Marginal Revenue Equals Marginal Cost at a Minimum
Consider a hypothetical firm with inverse demand \(p(q)=100-q\) and total cost \(TC(q)=120q-2q^2\), over \(0\le q\le20\). Find the stationary output and determine whether it maximizes profit.
Method. Apply the first-order condition, but do not stop there: use the second derivative and nearby profit values to classify the stationary point. The example avoids the mistake of treating \(MR=MC\) as sufficient evidence of a maximum.
Total revenue is \(TR(q)=100q-q^2\), so \(MR(q)=100-2q\). Marginal cost is \(MC(q)=120-4q\). Equating them gives \(100-2q=120-4q\), hence \(2q=20\) and \(q^*=10\). This is an interior feasible output.
Profit is \(\pi(q)=TR(q)-TC(q)=100q-q^2-(120q-2q^2)=q^2-20q\). Its second derivative is \(\pi''(q)=2>0\), so the stationary point is a strict local minimum, not a maximum. At \(q=10\), profit is \(100-200=-100\). As a nearby check, \(\pi(9)=81-180=-99\) and \(\pi(11)=121-220=-99\); both are greater than \(-100\).
The cost function has falling marginal cost over the stated interval: \(MC(q)=120-4q\), which remains positive from 120 at \(q=0\) to 40 at \(q=20\). The example shows that the equality locates a stationary point even when its curvature is wrong for profit maximization. Check: at \(q=10\), \(MR=100-20=80\) and \(MC=120-40=80\).
When the Interior Condition Does Not Apply
The first-order condition relies on the ability to vary output both upward and downward near the candidate. At a boundary, that assumption fails. A maximum can occur at a feasible endpoint even when marginal revenue and marginal cost are unequal there. The proper procedure is to check feasible stationary points and boundaries, rather than forcing an interior equality.
Worked Example: A Boundary Choice With No Feasible Stationary Point
A firm faces inverse demand \(p(q)=30-q\), has total cost \(TC(q)=40q\), and can choose output only in \(0\le q\le30\). Find the profit-maximizing feasible output.
Method. Check whether the marginal-equality equation has a feasible solution. If it does not, examine profit over the feasible interval and compare its boundaries. This avoids accepting an infeasible algebraic answer as the firm’s choice.
Total revenue is \(TR(q)=30q-q^2\), so \(MR(q)=30-2q\). Marginal cost is \(MC(q)=40\). Setting them equal gives \(30-2q=40\), or \(q=-5\), which is outside the feasible interval.
Profit is \(\pi(q)=TR(q)-TC(q)=30q-q^2-40q=-10q-q^2\). Its derivative is \(\pi'(q)=-10-2q<0\) throughout \(0\le q\le30\); profit is strictly decreasing on the feasible interval. At the boundaries, \(\pi(0)=0\) and \(\pi(30)=-300-900=-1{,}200\). Therefore the maximum over feasible outputs is at \(q=0\), not at the infeasible stationary solution \(q=-5\).
Check: because profit falls at every feasible increase in output, no positive feasible output can outperform zero output.
Exam Traps and a Reliable Decision Process
A common error is to stop after solving \(MR=MC\), or to replace marginal revenue with price. Under downward-sloping monopoly demand, selling an additional unit generally requires a price reduction that affects revenue on earlier units as well. Therefore, in general, \(MR\ne p\). The correct first-order comparison uses the derivative of total revenue, not the price alone.
Another trap is to infer a maximum from the equality without checking curvature or feasibility. A disciplined solution separates the candidate-finding step from the classification step:
Use \(\pi(q)=TR(q)-TC(q)\), and state which outputs are feasible. Confirm that the candidate will be interior before relying on an interior condition.
Differentiate to obtain \(\pi'(q)=MR(q)-MC(q)\), then solve \(MR(q)=MC(q)\). Reject any solution outside the feasible set.
Use \(\pi''(q^*)<0\) to establish a strict local maximum, if applicable. Check boundaries and other candidates when a global maximum is required; if \(\pi''(q^*)=0\), use another method.
If the second derivative is zero, the result is genuinely unresolved by the second-order test. For example, the functions \(\pi(q)=-(q-4)^4\) and \(\pi(q)=(q-4)^4\) both have first and second derivatives equal to zero at \(q=4\). The first has a local maximum there, while the second has a local minimum. Higher-order behavior or direct comparisons on either side are needed.
Check Your Understanding
For each problem, identify the relevant stationary point or boundary choice and apply the appropriate check.
- A firm has inverse demand \(p(q)=72-2q\) and total cost \(TC(q)=12q+q^2\), with \(0\le q\le36\). Find the profit-maximizing output, its price, and profit; verify the second-order condition. (Answer: \(MR=72-4q\), \(MC=12+2q\), \(q^*=10\), \(p=52\), \(\pi=300\), and \(\pi''=-6<0\).)
- Total revenue is \(TR(q)=90q-3q^2\) and total cost is \(TC(q)=18q+\frac{3}{2}q^2\), for \(0\le q\le15\). Find the stationary output and classify it. (Answer: \(MR=90-6q\), \(MC=18+3q\), \(q^*=8\), and \(\pi''=-9<0\), so it is a strict local maximum.)
- A firm has \(TR(q)=64q-2q^2\) and \(TC(q)=100+16q+q^2\), with \(0\le q\le32\). Find its stationary output and profit there, and explain whether the fixed cost changes the marginal-equality condition. (Answer: \(MR=64-4q\), \(MC=16+2q\), \(q^*=8\), \(\pi=384-(100+128+64)=92\), and \(\pi''=-6<0\); the fixed cost does not affect \(MC\) or the condition.)
- Inverse demand is \(p(q)=24-q\), marginal cost is constant at 30, and feasible output is \(0\le q\le24\). Determine the profit-maximizing output, checking whether the marginal-equality solution is feasible. (Answer: \(MR=24-2q\); equality gives \(q=-3\), outside the feasible interval; \(\pi(q)=-6q-q^2\) decreases on the interval, so the maximum is \(q=0\) with profit 0.)
- On a feasible interval around \(q=4\), compare \(\pi_1(q)=-(q-4)^4\) with \(\pi_2(q)=(q-4)^4\). Evaluate the first and second derivatives at 4 and classify each point. (Answer: both functions have \(\pi'(4)=0\) and \(\pi''(4)=0\); \(q=4\) is a maximum for \(\pi_1\) and a minimum for \(\pi_2\), so the zero-second-derivative test is inconclusive.)