Why a Stationary Point Needs a Second Test
For a firm with revenue \(R(q)\) and cost \(C(q)\), profit at output \(q\) is \(\pi(q)=R(q)-C(q)\). An interior output that satisfies marginal revenue equals marginal cost is a candidate for maximum profit, but that equality alone does not say whether profit is locally highest or lowest there. A second test examines how profit changes around the candidate.
This tutorial considers an output level strictly inside an interval on which revenue and cost are differentiable. The analysis classifies interior stationary points; it does not determine which endpoint of a feasible output interval is best. The central question is whether profit bends downward, bends upward, or has insufficient curvature information at the candidate.
The Second-Derivative Test
Assume \(R\) and \(C\) are twice differentiable near an interior candidate \(q^*\). Profit is \(\pi(q)=R(q)-C(q)\), so its first derivative is \(\pi'(q)=R'(q)-C'(q)\). At an interior stationary point, \(\pi'(q^*)=0\), equivalently \(MR(q^*)=MC(q^*)\), where \(MR=R'\) and \(MC=C'\).
The sign conditions are sufficient for the classifications, but a zero second derivative does not establish either one. At a local maximum of a twice-differentiable function, the second derivative cannot be positive; however, a zero value can occur at a maximum, a minimum, or a stationary point that is neither. In particular, do not treat \(\pi''=0\) as evidence that profit is flat over a whole interval.
Why the Test Works
The first derivative gives the direction in which profit is changing. At a stationary point it is zero, so the second derivative tells us how the slope changes nearby. If the slope is decreasing through zero, it is positive just before the candidate and negative just after it: profit rises and then falls. If the slope is increasing through zero, profit falls and then rises.
For a twice-differentiable function, this reasoning is captured locally by the second-order Taylor expansion around \(q^*\):
Here \(h\) is a small change in output and \(o(h^2)\) denotes a remainder that becomes negligible relative to \(h^2\) as \(h\) approaches zero. At a stationary point, \(\pi'(q^*)=0\), leaving the leading change \(\frac{1}{2}\pi''(q^*)h^2\). Since \(h^2\) is positive for either direction of a nonzero small change, a negative second derivative makes nearby profit lower, while a positive second derivative makes nearby profit higher. If \(\pi''(q^*)=0\), this term provides no classification.
Reading the Test Through Marginal Revenue and Marginal Cost
Because \(\pi'=MR-MC\), differentiating once more gives \(\pi''=MR'-MC'\), provided these derivatives exist. At a stationary point, marginal revenue and marginal cost are equal; whether that point is a maximum depends on how their slopes compare.
If \(MR'(q^*)<MC'(q^*)\), profit is locally concave at the candidate and the second-derivative test identifies a strict local maximum. If \(MR'(q^*)>MC'(q^*)\), the candidate is a strict local minimum. Equal slopes make the second derivative zero, so this test alone cannot decide. It is the relative slopes that matter, not whether marginal revenue slopes downward or marginal cost slopes upward in isolation.
Set \(\pi'(q)=0\), or equivalently \(MR(q)=MC(q)\), and solve for output.
Find \(\pi''(q)\), or calculate \(MR'(q)-MC'(q)\).
A negative value gives a strict local maximum; a positive value gives a strict local minimum. A zero value requires another test, such as checking the sign of \(\pi'\) on either side.
Worked Example: Confirming an Interior Profit Maximum
A firm has revenue \(R(q)=100q-q^2\) and cost \(C(q)=20+20q+q^2\), with output \(q>0\). Find and classify the interior stationary output, and calculate its profit. Method: construct profit, solve its first-order condition, and apply the second-derivative test.
Step 1: Construct the profit function.
Step 2: Find the stationary output. The first derivative is \(\pi'(q)=80-4q\). Set it equal to zero: \(80-4q=0\), so \(q^*=20\). This is an interior output because it is positive.
Step 3: Test its curvature. The second derivative is \(\pi''(q)=-4\), so \(\pi''(20)=-4<0\). Therefore \(q=20\) is a strict local maximum.
The marginal measures give the same result. \(MR(q)=100-2q\) and \(MC(q)=20+2q\); at \(q=20\), both equal \(60\). Their slopes are \(MR'=-2\) and \(MC'=2\), so \(MR'-MC'=-2-2=-4\), matching \(\pi''(20)\).
Step 4: Calculate and check profit. Revenue at \(q=20\) is \(R(20)=100(20)-20^2=2{,}000-400=1{,}600\). Cost is \(C(20)=20+20(20)+20^2=20+400+400=820\). Thus profit is \(1{,}600-820=780\). Direct substitution checks this: \(\pi(20)=80(20)-2(20^2)-20=1{,}600-800-20=780\).
The negative curvature confirms a local maximum, not by itself a comparison with any boundary output. The common mistake is to stop after finding \(MR=MC\); that equality finds a stationary candidate but does not classify it.
Different Stationary Points in One Profit Function
A profit function can have more than one stationary point, and the same test must be applied separately at each one. A candidate with positive curvature is a minimum, not a maximum, even though it also satisfies the first-order condition.
Worked Example: Classifying Two Stationary Outputs
A firm’s revenue and cost functions are \(R(q)=100q\) and \(C(q)=q^3-12q^2+136q+100\), for \(q>0\). Find all stationary outputs and classify each one. Method: factor the first derivative, then evaluate the second derivative at every root.
Step 1: Form profit and differentiate.
The stationary outputs are \(q=2\) and \(q=6\), both inside the stated positive-output domain.
Step 2: Classify \(q=2\). The second derivative is \(\pi''(q)=-6q+24\), so \(\pi''(2)=-12+24=12>0\). Thus \(q=2\) is a strict local minimum. As a check on the marginal slopes, \(MR=100\), \(MC=3q^2-24q+136\), and \(MC(2)=12-48+136=100\). Also \(MR'=0\), \(MC'(2)=6(2)-24=-12\), and \(MR'-MC'=12\).
Step 3: Classify \(q=6\). Here \(\pi''(6)=-36+24=-12<0\), so \(q=6\) is a strict local maximum. Again, \(MR=100\) and \(MC(6)=108-144+136=100\). The slope difference is \(MR'-MC'=0-(36-24)=-12\), agreeing with the second derivative.
Step 4: Check the profit values. At \(q=2\), \(\pi(2)=-8+48-72-100=-132\). At \(q=6\), \(\pi(6)=-216+432-216-100=-100\). These values do not replace the curvature test, but are consistent with the two distinct stationary classifications. The common mistake is to call both points maxima just because both satisfy \(MR=MC\).
When the Second-Derivative Test Is Inconclusive
A zero second derivative does not mean that profit has no local maximum. For example, if \(\pi(q)=50-(q-4)^4\), then \(\pi'(4)=0\) and \(\pi''(4)=0\), yet \((q-4)^4>0\) for every \(q\ne4\), so \(\pi(q)<\pi(4)\) nearby. Output \(4\) is a strict local maximum, found by inspecting the profit function beyond its second derivative.
A zero second derivative can also occur at a point that is neither a maximum nor a minimum. In that case, checking the first derivative on both sides, or comparing nearby profit values, can resolve the classification.
Worked Example: A Stationary Point That Is Not an Extremum
A firm’s revenue is \(R(q)=100q\), and its cost is \(C(q)=100q-30-(q-5)^3\), over a neighborhood of \(q=5\). Classify the stationary output at \(q=5\). Method: when the second derivative vanishes, test the sign of the first derivative on either side.
Step 1: Calculate profit and its derivatives.
At \(q=5\), \(\pi'(5)=0\) and \(\pi''(5)=0\). Therefore the second-derivative test is inconclusive. The marginal condition holds as well: \(MR=100\), while \(MC(q)=100-3(q-5)^2\), so \(MC(5)=100\).
Step 2: Check the sign of the first derivative around \(5\). For every \(q\ne5\), \(\pi'(q)=3(q-5)^2>0\), on both sides of \(5\). Profit is increasing through the stationary point, rather than increasing before it and decreasing after it or the reverse.
Step 3: Verify with nearby values. \(\pi(4)=30+(-1)^3=29\), \(\pi(5)=30\), and \(\pi(6)=30+1^3=31\). The value rises through \(q=5\), so \(q=5\) is neither a local maximum nor a local minimum. The common mistake is to infer that every stationary point is an extremum; a zero first derivative only identifies a candidate.
Concavity and a Unique Interior Maximum
The second derivative can classify a point locally, or it can establish a conclusion over an entire interval. If \(\pi''(q)<0\) throughout an interval, profit is strictly concave there: \(\pi'(q)\) is strictly decreasing. Consequently, there can be at most one stationary point in that interval. If such a point exists, the decreasing slope is positive to its left and negative to its right, making it the unique maximum within the interval.
More generally, if \(\pi''(q)\leq0\) throughout an interval, profit is concave there, and any interior stationary point is a global maximum on that interval. Strict concavity makes that maximum unique. These conclusions require the stated curvature condition across the interval, not just at one output. A negative second derivative at one point supports a strict local maximum; it does not, on its own, prove global optimality.
Common Mistakes
- Treating \(MR=MC\) as sufficient for a maximum. It is the stationary-point condition. A positive second derivative identifies a minimum instead.
- Reversing the slope comparison. Since \(\pi''=MR'-MC'\), a maximum requires \(MR'<MC'\), not \(MR'>MC'\).
- Calling a zero second derivative a maximum or a minimum. The test is inconclusive at zero; inspect the first derivative around the point or compare nearby profit values.
- Using a local result as a global one. Curvature at a single point classifies that point locally. A global conclusion from concavity requires the curvature condition over the relevant interval.
Check Your Understanding
For each problem, classify the stated interior candidate using the appropriate derivative test. Work through the calculation before checking the answer.
- A firm has profit \(\pi(q)=84q-2q^2-12\). Find its stationary output, classify it, and calculate profit there. (Answer: \(q=21\), strict local maximum, profit \(870\).)
- Revenue is \(R(q)=90q-q^2\), and cost is \(C(q)=10+18q+q^2\). Find the stationary output, verify \(MR=MC\), and classify it. (Answer: \(q=18\); \(MR=MC=54\); \(\pi''=-4\), so it is a strict local maximum.)
- Profit is \(\pi(q)=-q^3+9q^2-18q+5\). Find both stationary outputs and classify each. (Answer: \(q=3-\sqrt{3}\) is a strict local minimum; \(q=3+\sqrt{3}\) is a strict local maximum.)
- Profit near \(q=3\) is \(\pi(q)=12+(q-3)^3\). The first and second derivatives both vanish at \(q=3\). Use the first derivative on either side to classify it. (Answer: neither a local maximum nor a local minimum; \(\pi'(q)=3(q-3)^2>0\) on both sides.)
- On the interval \(0<q<30\), profit is \(\pi(q)=72q-2q^2-100\). Show that its stationary output is the unique maximum on this interval, and calculate the maximum profit. (Answer: \(\pi''=-4<0\) throughout the interval; the unique stationary output is \(q=18\), with profit \(548\).)