Difficulty: Intermediate. Prerequisites: Limits and Continuity (Sections 2.1–2.5), basic derivative concept.
Revenue is what comes in, cost is what goes out, and profit is the difference. The marginal version of each function is its derivative, and it estimates how much the next single unit adds to revenue, cost, or profit. Most exam problems ask you to build these functions from a word problem, differentiate, and interpret the result at a specific production level.
Revenue function, R(x)
The total income from selling x units. If each unit sells at a fixed price p, then R(x) = p · x. When price depends on demand (a price-demand equation), R(x) = p(x) · x.
In simple terms: total money coming in from sales.
Cost function, C(x)
The total cost of producing x units. Typically includes a fixed cost (constant term) plus variable costs that depend on x.
In simple terms: everything the business spends to make x items.
Profit function, P(x)
P(x) = R(x) − C(x). Revenue minus cost. A negative profit is a loss.
Think of it as: what is left after paying all production costs.
Marginal cost, C′(x)
The derivative of the cost function. Estimates the cost of producing one additional unit beyond x units.
In simple terms: roughly how much the next unit costs to make.
Marginal revenue, R′(x)
The derivative of the revenue function. Estimates the additional revenue from selling one more unit beyond x units.
Marginal profit, P′(x)
The derivative of the profit function. Estimates the change in profit from producing and selling one more unit beyond x units.
Average cost, C̄(x)
C̄(x) = C(x)/x. The cost per unit when x units are produced.
Think of it as: spreading total cost evenly across all units made.
Average profit, P̄(x)
P̄(x) = P(x)/x. The profit per unit when x units are produced and sold.
Marginal average cost, C̄′(x)
The derivative of the average cost function. Estimates how the per-unit cost changes when one more unit is produced.
Actual cost of the (x+1)th unit
C(x+1) − C(x). The exact (not estimated) additional cost. Exam questions often ask you to compare this to the marginal cost estimate C′(x).
Most exam questions give you a scenario and ask you to identify or construct these functions.
Revenue from a fixed price: If an oil company sells barrels at $100 each, R(x) = 100x.
Revenue from a price-demand equation: If the demand equation is x = 9000 − 30p, solve for p first: p = (9000 − x)/30 = 300 − x/30. Then R(x) = x · p = 300x − x²/30.
Cost is usually given directly: e.g. C(x) = 0.1x³ − 2x² + 10000.
Profit is always R(x) − C(x): Subtract the entire cost function from the revenue function. Watch signs carefully when distributing the negative.
The marginal function is simply the derivative.
Given P(x) = 15x − 0.05x² − 600, the marginal profit is P′(x) = 15 − 0.1x.
Given C(x) = 0.005x³ − 2x + 5000, the marginal cost is C′(x) = 0.015x² − 2.
Given C(x) = x² − 5x + 1000, the average cost is C̄(x) = x − 5 + 1000/x, and the marginal average cost is C̄′(x) = 1 − 1000/x².
This is the most common exam application. The pattern:
Find the marginal function (take the derivative).
Evaluate it at the current production level.
Interpret: this value approximates the change from producing/selling one additional unit.
Example: A manufacturer has cost function C(x) = 0.1x³ − 2x² + 10000. Estimate the cost of the 47th toy.
C′(x) = 0.3x² − 4x.
C′(46) = 0.3(46)² − 4(46) = 0.3(2116) − 184 = 634.8 − 184 = $450.80.
Interpretation: producing the 47th toy costs approximately $450.80.
For comparison, the actual cost of the 47th toy is C(47) − C(46). The marginal estimate is close but not exact.
A negative profit means a loss. When asked "does the result make sense?", check whether the production level is too low to cover fixed costs, or whether costs are growing faster than revenue.
Example: P(x) = 2x² − 59x − 72000. At x = 102: P(102) = 2(10404) − 59(102) − 72000 = 20808 − 6018 − 72000 = −57210. A loss of $57,210 at only 102 units may make sense if fixed costs are high and the break-even point is much larger.
P̄(x) = P(x)/x. At x = 50 mowers with P(x) = 30x − 0.03x² − 750:
P̄(50) = P(50)/50 = (1500 − 75 − 750)/50 = 675/50 = $13.50 per mower.
The marginal average profit P̄′(x) tells you whether that per-unit figure is rising or falling as you produce more.
P(x) = R(x) - C(x)R(x) = p \cdot x \quad \text{(fixed price)} \qquad R(x) = p(x) \cdot x \quad \text{(variable price)}\bar{C}(x) = \frac{C(x)}{x} \qquad \bar{P}(x) = \frac{P(x)}{x}\text{Marginal cost} = C'(x) \qquad \text{Marginal revenue} = R'(x) \qquad \text{Marginal profit} = P'(x)\text{Actual cost of unit } (x+1) = C(x+1) - C(x) \approx C'(x)The marginal value at x estimates the exact change from unit x to unit x + 1. The approximation improves when the function is roughly linear over that interval.
Marginal analysis is how businesses decide whether to produce one more unit. If the marginal revenue of the next unit exceeds its marginal cost, that unit adds to profit and is worth making. This is the economic principle behind production-level decisions in manufacturing, retail pricing, and service capacity planning.
"Marginal cost IS the cost of the next unit." Close, but not exact. Marginal cost is the derivative, which approximates the cost of the next unit. The actual cost of the (x+1)th unit is C(x+1) − C(x), which is slightly different.
"A negative profit means I did the maths wrong." Not necessarily. Low production volumes often produce negative profit because fixed costs have not been covered yet. Always check whether the result is plausible given the context.
"Marginal profit at x = 20 means the profit on the 20th unit." It estimates the additional profit from producing unit 21 (the next unit beyond 20), not unit 20 itself.
Forgetting to subtract the entire cost function when building P(x). Distribute the negative sign across every term of C(x). A common arithmetic error is to negate only the first term.
⚠️ "Use marginal analysis to estimate..." is the signature phrasing. It means: take the derivative and evaluate at the given x. Do not compute the actual difference.
⚠️ Building P(x) from R(x) and C(x) is tested in nearly every variant of this exam. Practise the subtraction carefully.
⚠️ Average cost and marginal average cost questions are less common but do appear. Know C̄(x) = C(x)/x and how to differentiate it (quotient rule or rewrite as a sum).
⚠️ When a problem says "does the result make sense?", give a one-sentence interpretation (e.g. the company is operating at a loss because production volume is too low to cover fixed costs).
True or false: P(x) = C(x) − R(x). False. P(x) = R(x) − C(x).
Fill in the blank: The marginal cost C′(x) estimates the cost of the ______ unit. Next (the (x+1)th unit).
True or false: If P(100) = −50000, the calculation must contain an error. False. A loss at low production can be correct.
Fill in the blank: Average cost is C̄(x) = ______. C(x)/x.
True or false: Marginal revenue R′(x) gives the exact revenue from the next unit sold. False. It is an approximation. The exact change is R(x+1) − R(x).
Q: A company sells barrels at $100 each. Cost is C(x) = 0.005x³ − 2x + 5000. What is the profit function?
A: R(x) = 100x. P(x) = R(x) − C(x) = 100x − 0.005x³ + 2x − 5000 = −0.005x³ + 102x − 5000.
Q: Using the profit function above, find the marginal profit and use it to estimate the profit from selling the 21st barrel.
A: P′(x) = −0.015x² + 102. P′(20) = −0.015(400) + 102 = −6 + 102 = $96. The 21st barrel adds approximately $96 to profit.
Q: A toy manufacturer has C(x) = 0.1x³ − 2x² + 10000. What is the actual cost of the 47th toy? What is the marginal estimate?
A: Actual = C(47) − C(46). C(47) = 0.1(103823) − 2(2209) + 10000 = 10382.3 − 4418 + 10000 = 15964.3. C(46) = 0.1(97336) − 2(2116) + 10000 = 9733.6 − 4232 + 10000 = 15501.6. Actual = $462.70. Marginal: C′(x) = 0.3x² − 4x. C′(46) = 0.3(2116) − 184 = $450.80. Close but not identical.
Q: The price-demand equation is x = 9000 − 30p and C(x) = 150000 + 30x. Find the revenue function R(x) and the marginal profit function.
A: Solve for p: p = 300 − x/30. R(x) = x(300 − x/30) = 300x − x²/30. P(x) = R(x) − C(x) = 300x − x²/30 − 150000 − 30x = 270x − x²/30 − 150000. P′(x) = 270 − x/15.
Q: Find the average profit per mower if P(x) = 30x − 0.03x² − 750 and 50 mowers are produced.
A: P(50) = 1500 − 75 − 750 = 675. P̄(50) = 675/50 = $13.50 per mower.
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Connections to other topics: Marginal analysis is the first real application of derivatives in this course. The derivative rules from Sections 3.1–3.3 (power rule, product rule, quotient rule) are what you use to compute marginal functions efficiently. Optimisation problems later in the course (finding the production level that maximises profit) build directly on setting P′(x) = 0.