9 min readConcept sheet10-mark problem: maximum revenue or profit without calculus

Quadratic models: revenue, profit and the vertex

When price falls as sales rise, revenue and profit become quadratics (parabolas). The top of the parabola, the vertex at x = −b/2a, gives the best output.

Shapea < 0: opens down, has a maximum

Vertexx = −b ÷ 2a, then find y

RevenueR = p × q with p = a − bq

Break-evenRoots of P(x) = 0

Quadratic model formulas

General formy = ax² + bx + c

Vertex x-coordinatex = −b ÷ 2a

Vertex y-coordinateSubstitute x back into the function

Revenue from linear demand p = a − bqR(q) = aq − bq²

Maximum revenue (linear demand)At q = a ÷ 2b

ProfitP(q) = R(q) − C(q)

Break-even pointsSolve P(q) = 0 (two roots)

Sketching a parabola

Maximum revenue

Problem

A Colombo three-wheeler hire firm faces demand p = 600 − 2q (p in Rs. per trip, q trips a day). Find the revenue function and the price that maximises revenue.

Working

R(q) = p × q = (600 − 2q)q = 600q − 2q².
a = −2 < 0, so the parabola has a maximum at the vertex.
q = −b ÷ 2a = −600 ÷ (2 × −2) = 150 trips.
Price p = 600 − 2(150) = 300.
R(150) = 600 × 150 − 2 × 150² = 90,000 − 45,000 = 45,000.

Answer

Charge Rs. 300 a trip for 150 trips; maximum revenue Rs. 45,000 a day.

Maximum profit and break-even

Problem

For the same firm, cost is C(q) = 100q + 20,000. Find the profit function, the profit-maximising output and price, the maximum profit, and the break-even outputs.

Working

P(q) = 600q − 2q² − 100q − 20,000 = −2q² + 500q − 20,000.
Vertex: q = −500 ÷ (2 × −2) = 125. Price = 600 − 2(125) = 350.
P(125) = −2(15,625) + 62,500 − 20,000 = −31,250 + 62,500 − 20,000 = 11,250.
Break-even: −2q² + 500q − 20,000 = 0, so q² − 250q + 10,000 = 0.
D = 62,500 − 40,000 = 22,500, √D = 150, q = (250 ± 150) ÷ 2 = 200 or 50.
Note: profit is greatest at 125 trips, not at the revenue maximum of 150.

Answer

Best output 125 trips at Rs. 350; maximum profit Rs. 11,250; break-even at 50 and 200 trips.

Equilibrium with quadratic curves

Problem

Demand is p = 100 − q² and supply is p = q² + 2q + 40. Find the market equilibrium.

Working

Set equal: 100 − q² = q² + 2q + 40.
0 = 2q² + 2q − 60, so q² + q − 30 = 0.
(q + 6)(q − 5) = 0, so q = 5 (q = −6 makes no sense).
p = 100 − 5² = 75. Check with supply: 25 + 10 + 40 = 75. ✓

Answer

q = 5 at p = 75.

In Sri Lankan business

A Sri Lankan three-wheeler hire firm or hotel that cuts prices to win customers sees revenue rise, peak, then fall. The vertex of the revenue parabola shows the price that brings the most revenue; the profit vertex shows the best price once costs count.

Exam angle

Short answer: “Find the vertex x-coordinate of y = −3x² + 24x + 5.” x = −24 ÷ (2 × −3) = 4.

Structured question: “Demand p = 80 − q. Find R(q) and the maximum revenue.” R = 80q − q²; vertex q = 40; R = 3,200 − 1,600 = 1,600.

Essay: “Demand p = 600 − 2q, cost C = 100q + 20,000. Find the maximum profit, its price, and the break-even outputs (without calculus).” P = −2q² + 500q − 20,000; q = 125, p = 350, P = 11,250; break-even 50 and 200.

Remember it

Vertex = −b over 2a. Revenue peaks where the demand line's quantity is half its maximum; profit peaks a little earlier because costs rise with output.