Mathematical Economics · Optimisation
Profit Maximisation with Calculus: First and Second Order Conditions
Setting the derivative to zero finds a flat point. Flat points can be peaks or valleys. The second-order condition is what tells you which one you have found — and skipping it can reverse your answer entirely.
Here is a question that catches out a lot of otherwise competent students.
You are given a profit function. You differentiate it, set the derivative to zero, solve, and write down your answer. Standard procedure. But your derivative was a quadratic, so it gave you two solutions.
Which one is the profit maximum?
You cannot tell from the first derivative. Both are points where the slope is flat, and a flat slope is exactly as consistent with the bottom of a valley as with the top of a hill. One of your two answers maximises profit. The other minimises it. Reporting the wrong one is not a small error — it is advising a firm to adopt its worst possible output level.
The tool that separates them is the second-order condition, and this post explains what it is, why it works, and how to write it up so you get the marks.
The first-order condition: necessary, not sufficient
Start with what you already know. Profit is revenue minus cost, and to find a turning point you differentiate and set the result to zero.
π(Q) = TR(Q) − TC(Q)
dπ/dQ = 0, which gives MR = MC
Key term
First-order condition (FOC) — the requirement that the first derivative equals zero at an optimum. It identifies stationary points: places where the function is momentarily flat. Every interior maximum satisfies it, but so does every interior minimum, and so does every point of inflection.
That last sentence is the whole issue. The FOC is necessary but not sufficient. It narrows your candidates down to a short list. It does not tell you which candidate you want.
Think about what a flat slope actually means. At the summit of a hill, the ground is level. At the bottom of a bowl, the ground is also level. If all you know is “the ground here is level,” you cannot say whether you are at the top or the bottom. You need to know about the curvature — and curvature is what the second derivative measures.
The second-order condition: reading the curvature
The rule
At a stationary point, evaluate the second derivative:
• If d²π/dQ² < 0 → the function is concave (curving downwards) → maximum
• If d²π/dQ² > 0 → the function is convex (curving upwards) → minimum
• If d²π/dQ² = 0 → inconclusive; you need further tests
• If d²π/dQ² > 0 → the function is convex (curving upwards) → minimum
• If d²π/dQ² = 0 → inconclusive; you need further tests
A memory aid that actually works: a negative second derivative means the slope is falling as you move right. It was positive before the peak and negative after it. A falling slope is a hilltop. Sad face, curving down, maximum.
There is an economic reading of this too, and it is worth knowing because examiners like it. The second derivative of profit is:
d²π/dQ² = dMR/dQ − dMC/dQ
So the condition d²π/dQ² < 0 is really the statement that marginal cost must be rising faster than marginal revenue at the optimum. In the ordinary case where MR is flat or falling and MC is rising, this is automatically satisfied — which is why the SOC feels invisible in simple textbook problems. It only bites when cost structures are unusual, and that is precisely when exam questions test it.
A worked example where it matters
Take a firm with the profit function:
π(Q) = −Q³ + 9Q² − 15Q − 10
Step 1 — Apply the first-order condition.
dπ/dQ = −3Q² + 18Q − 15 = 0
Divide through by −3 to simplify:
Q² − 6Q + 5 = 0, so (Q − 1)(Q − 5) = 0
Two candidates: Q = 1 and Q = 5. The FOC has taken us as far as it can. Both are stationary points; it is silent on which is which.
Step 2 — Apply the second-order condition. Differentiate again:
d²π/dQ² = −6Q + 18
Now evaluate at each candidate:
| Candidate | Second derivative | Sign | Verdict |
|---|---|---|---|
| Q = 1 | −6(1) + 18 = +12 | Positive | Minimum |
| Q = 5 | −6(5) + 18 = −12 | Negative | Maximum |
Step 3 — Confirm with the actual profit values. This is a useful sanity check and takes seconds.
π(1) = −1 + 9 − 15 − 10 = −17
π(5) = −125 + 225 − 75 − 10 = +15
π(5) = −125 + 225 − 75 − 10 = +15
The second-order condition was right. Q = 5 yields a profit of £15; Q = 1 yields a loss of £17. A student who solved the FOC, took the first root they found, and reported Q = 1 would have recommended the firm’s worst output choice — a swing of £32 in the wrong direction.
Common error
Stopping at the first-order condition and assuming any stationary point is the maximum. In simple quadratic problems this happens to work, which breeds a bad habit. The moment a question gives you a cubic profit or cost function — and examiners use cubics precisely for this reason — the FOC produces two roots and the habit fails. Always compute the second derivative, always state its sign, always name the verdict.
What to write to get full marks
Markers award method marks for the structure, not just the number. Use this sequence explicitly:
1. State the objective function.
2. “Applying the first-order condition, dπ/dQ = 0.” Differentiate and solve.
3. List all stationary points found. Do not silently discard one.
4. “Applying the second-order condition, d²π/dQ² = …” Evaluate at each candidate.
5. State the sign and the conclusion: “Since d²π/dQ² < 0 at Q = 5, this is a maximum.”
6. Check the economics: is the answer non-negative? Is profit positive? Should the firm shut down?
2. “Applying the first-order condition, dπ/dQ = 0.” Differentiate and solve.
3. List all stationary points found. Do not silently discard one.
4. “Applying the second-order condition, d²π/dQ² = …” Evaluate at each candidate.
5. State the sign and the conclusion: “Since d²π/dQ² < 0 at Q = 5, this is a maximum.”
6. Check the economics: is the answer non-negative? Is profit positive? Should the firm shut down?
Step six matters more than students expect
Calculus finds mathematical optima; it knows nothing about economics. If your algebra returns Q* = −4, the mathematics is fine and the answer is meaningless, because a firm cannot produce negative output. The economically relevant answer is the corner solution Q = 0. Similarly, if profit at your optimum is more negative than fixed costs, the firm does better shutting down. Always ask whether the number your calculus produced makes economic sense before you underline it.
Extending to more than one variable
Once a firm chooses two things at once — two products, or labour and capital — the same logic extends, with more bookkeeping.
The first-order conditions become a set of partial derivatives, all set to zero simultaneously. For profit depending on two outputs Q₁ and Q₂:
∂π/∂Q₁ = 0 and ∂π/∂Q₂ = 0
You then solve those as simultaneous equations. The second-order condition becomes a condition on the matrix of second derivatives — the Hessian — requiring, for a maximum, that ∂²π/∂Q₁² and ∂²π/∂Q₂² are both negative and that the determinant condition holds. Most undergraduate courses state this rather than derive it, and it is enough to know that the underlying idea is unchanged: the first derivatives locate flat points, the second derivatives establish curvature.
Practice questions
Q1. A competitive firm faces a market price of £66 and has TC = 100 + 6Q + 0.5Q². Find the profit-maximising output, verify it with the SOC, and compute profit.
Q2. A firm has π(Q) = −2Q³ + 30Q² − 96Q − 20. Find both stationary points and classify each.
Q3. Explain in economic language what d²π/dQ² < 0 requires about the behaviour of MR and MC.
Q4. A student solves an FOC, obtains Q* = −3, and reports it as the profit-maximising output. What has gone wrong and what is the correct answer?
Worked answers
A1. Profit is π = 66Q − (100 + 6Q + 0.5Q²) = −100 + 60Q − 0.5Q².
FOC: dπ/dQ = 60 − Q = 0, so Q* = 60.
SOC: d²π/dQ² = −1, which is negative, confirming a maximum.
Profit: π = −100 + 60(60) − 0.5(3600) = −100 + 3,600 − 1,800 = £1,700. Note that here the second derivative is a constant, so the SOC holds everywhere — typical of quadratic profit functions, and the reason these problems feel like the SOC is a formality.
A2. FOC: dπ/dQ = −6Q² + 60Q − 96 = 0. Divide by −6: Q² − 10Q + 16 = 0, which factorises as (Q − 2)(Q − 8) = 0. Candidates: Q = 2 and Q = 8.
SOC: d²π/dQ² = −12Q + 60. At Q = 2 this is +36 (positive, so a minimum). At Q = 8 it is −36 (negative, so a maximum).
Checking: π(2) = −16 + 120 − 192 − 20 = −108, while π(8) = −1,024 + 1,920 − 768 − 20 = +108. The classification is confirmed.
A3. Since d²π/dQ² = dMR/dQ − dMC/dQ, requiring it to be negative means dMC/dQ > dMR/dQ — marginal cost must be rising more steeply than marginal revenue at the chosen output. Intuitively, MC must be cutting MR from below. If MC crossed MR from above, then just past the crossing point MR would exceed MC and the firm could raise profit by expanding — so that crossing would be a profit minimum, not a maximum.
A4. The algebra may be perfectly correct, but the answer violates a constraint the mathematics does not know about: output cannot be negative. The FOC located an unconstrained stationary point that lies outside the feasible region. The correct approach is to recognise this as a corner solution: since profit is decreasing across the entire feasible range Q ≥ 0, the best feasible choice is Q = 0 — the firm should produce nothing. This is exactly the shut-down case, and stating it explicitly is what earns the mark.
The short version
• FOC (first derivative = 0) finds stationary points. Necessary, not sufficient.
• SOC (sign of the second derivative) classifies them. Negative means maximum.
• Cubic profit functions give two candidates — one max, one min. The SOC is the only thing separating them.
• Economically, the SOC means MC rises faster than MR; MC must cut MR from below.
• Always sanity-check the economics: negative output means a corner solution at Q = 0.
• Write out both conditions explicitly. Markers award the method.
• SOC (sign of the second derivative) classifies them. Negative means maximum.
• Cubic profit functions give two candidates — one max, one min. The SOC is the only thing separating them.
• Economically, the SOC means MC rises faster than MR; MC must cut MR from below.
• Always sanity-check the economics: negative output means a corner solution at Q = 0.
• Write out both conditions explicitly. Markers award the method.
References
1. Chiang, A.C. & Wainwright, K. (2005) Fundamental Methods of Mathematical Economics. 4th edn. New York: McGraw-Hill.
2. Sydsaeter, K., Hammond, P. & Strom, A. (2016) Essential Mathematics for Economic Analysis. 5th edn. Harlow: Pearson.
3. Simon, C.P. & Blume, L. (1994) Mathematics for Economists. New York: W.W. Norton.
4. Nicholson, W. & Snyder, C. (2016) Microeconomic Theory: Basic Principles and Extensions. 12th edn. Boston: Cengage.
5. Varian, H.R. (2014) Intermediate Microeconomics: A Modern Approach. 9th edn. New York: W.W. Norton.
Take this further
Optimisation underpins every firm and consumer model you will meet. The full treatment, with worked exam questions and the diagrams that make curvature intuitive, is in Economics Made Simple: A Supplementary Guide for AP and Cambridge A-Level Students.
To see these conditions applied to real firms and markets, Economics Case Studies: 100 Real-World Examples works through the applications. Both come together in the Economics Made Simple bundle.