ECON5005 · Quantitative Tools for Economics
Single-Variable Optimisation
Single-variable optimisation is where the calculus of ECON5005 Quantitative Tools for Economics at the University of Sydney becomes decision-making: you choose one variable to make an objective as large (or small) as possible. This chapter builds the two-condition engine — the first-order condition f′(x) = 0 to locate a candidate and the second-order condition f″(x) to classify it — then applies it to the unit's headline case, profit maximisation with MR = MC and break-even.
What this chapter covers
- 01Set up an optimisation problem: objective function, choice variable, global vs local optima
- 02Apply the first-order condition f′(x*) = 0 to find stationary points
- 03Classify with the second-order condition: f″ < 0 for a max, f″ > 0 for a min
- 04Handle the f″ = 0 case with a sign chart of f′
- 05Build profit pi(Q) = TR - TC and derive the FOC MR = MC
- 06Confirm a profit maximum via the SOC pi″ < 0 (MC rising through MR)
- 07Specialise to a price-taking firm: MR = P, so the rule becomes P = MC
- 08Solve break-even, pi(Q) = 0, and reject economically invalid roots
Profit maximisation with MR = MC and break-even
- +1Inverse demand: P = 24 - Q, so total revenue TR = P.Q = (24 - Q)Q = 24Q - Q^2.
- +1Marginal revenue MR = d(TR)/dQ = 24 - 2Q (note MR falls twice as fast as demand).
- +1Marginal cost MC = d(TC)/dQ = 2Q.
- +1First-order condition MR = MC: 24 - 2Q = 2Q, so 24 = 4Q and Q* = 6.
- +1Second-order condition: pi(Q) = 24Q - Q^2 - (Q^2 + 40) = -2Q^2 + 24Q - 40, so pi'' = -4 < 0 -> a maximum.
- +1Price at the optimum: P* = 24 - 6 = 18.
- +1Maximum profit: pi(6) = -2(36) + 144 - 40 = 32 (check: TR = 108, TC = 76, pi = 32).
- +1Break-even pi(Q) = 0: -2Q^2 + 24Q - 40 = 0 -> Q^2 - 12Q + 20 = 0 -> (Q - 2)(Q - 10) = 0, so Q = 2 or Q = 10.
Key terms
- Objective function
- The quantity f(x) you are trying to maximise or minimise (for a firm, profit); x is the single choice variable you control.
- First-order condition (FOC)
- At an interior optimum the tangent is flat: f′(x*) = 0. It locates stationary points but does not, by itself, say whether they are maxima, minima or inflections.
- Second-order condition (SOC)
- The curvature test that classifies a stationary point: f″(x*) < 0 gives a local maximum, f″(x*) > 0 a local minimum; f″ = 0 is inconclusive.
- Stationary point
- Any point where f′(x) = 0. Maxima, minima and inflection points all qualify, which is why the SOC is needed to tell them apart.
- Marginal revenue (MR)
- The derivative of total revenue, MR = d(TR)/dQ. For a price-taker MR equals the fixed market price P.
- Marginal cost (MC)
- The derivative of total cost, MC = d(TC)/dQ — the added cost of one more unit of output.
- Profit maximisation (MR = MC)
- The FOC of pi(Q) = TR(Q) - TC(Q). Profit is greatest at the output where marginal revenue equals marginal cost, provided the SOC pi″ < 0 holds.
- Break-even quantity
- An output at which profit is zero, pi(Q) = 0 (total revenue equals total cost). A quadratic profit function typically has two of them, straddling the profit-maximising quantity.
Single-Variable Optimisation FAQ
Why isn't f′(x) = 0 enough to prove I have a maximum?
Because the first-order condition is satisfied at minima and inflection points too. Setting the derivative to zero only finds a candidate; you must then check the second-order condition — f″(x*) < 0 confirms a maximum, f″(x*) > 0 a minimum. If f″ = 0, build a sign chart of f′ (positive then negative around the point signals a maximum). In the exam, the SOC line usually carries its own mark.
What is the difference between the MR = MC rule and P = MC?
MR = MC is the general profit-maximising condition for any firm, found from pi′(Q) = 0. A perfectly competitive (price-taking) firm sells at a fixed market price P, so its marginal revenue is constant at MR = P, and the rule specialises to P = MC (with marginal cost rising). A firm with market power faces a downward-sloping demand, so its MR lies below price and the two rules give different quantities.
Can AI help me with single-variable optimisation in ECON5005?
Yes — used as a study aid. Sia, the AskSia AI tutor, can explain the method step by step: how to set up the objective, apply the first-order condition, run the second-order test, and interpret MR = MC and break-even, using practice problems with your own numbers so you learn the reasoning. It does not sit your quizzes, mid-semester or final exam or hand you assessment answers, and it cannot promise a grade; treat it as a tutor that helps you work each step yourself and confirm the exact assessment rules on Canvas.
Studying with AI? Sia — free AI economics tutor works through ECON5005 step by step.
Exam move
Optimisation is a method you drill, not facts you memorise. Rehearse the four-step recipe until it is automatic: differentiate and solve f′(x) = 0, reject any economically invalid root (no negative quantities or prices), classify with the second-order condition, then substitute back to report the optimal value. For firm problems always write TR and TC first, differentiate to MR and MC, solve MR = MC, and finish with the pi″ < 0 line plus price, profit and break-even. Because this reasoning anchors a long multi-part question on the final exam, practise past-style profit problems end to end under time (roughly 2 minutes per mark) rather than stopping at the quantity. Confirm the exact assessment weights, and the open/closed-book and calculator policy, on your Canvas page.
