Difficulty: Intermediate | Prerequisites: Basic derivatives (power rule), algebra of polynomials
This material sits at the intersection of basic calculus and business modelling. You are given (or asked to build) functions that describe what a company earns, what it spends, and what is left over. Then you use derivatives to estimate how those quantities change when one more unit is produced or sold. If you are comfortable with the power rule and basic polynomial algebra, you have the prerequisites. The payoff is that marginal analysis lets you approximate the effect of the "next unit" without recalculating the entire function, which is the core idea that links calculus to real economic decision-making.
Revenue is price times quantity, cost is given or modelled, and profit is revenue minus cost. The derivative of any of these (the "marginal" version) estimates the change from producing or selling one additional unit. Average functions divide the total by the number of units, and their derivatives tell you whether that per-unit figure is rising or falling.
Revenue function, R(x)
The total income from selling x units. Calculated as price per unit times quantity: R(x) = p * x. When price itself depends on demand (a price-demand equation), you substitute the expression for p into the product.
In simple terms, this is "all the money coming in before you subtract any costs."
Cost function, C(x)
The total expense of producing x units. Typically given in the problem as a polynomial. It includes both fixed costs (the constant term, which you pay even if you produce nothing) and variable costs (terms that depend on x).
Think of it as the bill you have to pay regardless of how much you earn.
Profit function, P(x)
What remains after costs are subtracted from revenue: P(x) = R(x) - C(x). Profit can be negative, which simply means costs exceed revenue at that production level.
In simple terms, this is "what you actually keep."
Marginal cost, C'(x)
The derivative of the cost function. It estimates how much the total cost increases when you produce one additional unit beyond x.
Think of it as the price tag on "just one more."
Marginal revenue, R'(x)
The derivative of the revenue function. It estimates the additional revenue earned from selling one more unit beyond x.
Marginal profit, P'(x)
The derivative of the profit function. It estimates the additional profit from selling one more unit beyond x. Equivalently, P'(x) = R'(x) - C'(x).
Average cost function, C-bar(x)
Total cost divided by the number of units: C-bar(x) = C(x) / x. It tells you the cost per unit at a given production level.
Average profit function, P-bar(x)
Total profit divided by the number of units: P-bar(x) = P(x) / x.
Marginal average cost, C-bar'(x)
The derivative of the average cost function. It tells you whether the per-unit cost is rising or falling as production increases.
Price-demand equation
A relationship between the price p and the number of units x that consumers will buy at that price. Often given as x = f(p), which you rearrange to p = g(x) before building the revenue function.
Identify the selling price. If the price is a fixed number (e.g. $100 per barrel), then R(x) = price * x.
If a price-demand equation is given (e.g. x = 9000 - 30p), isolate p first: p = (9000 - x) / 30 = -x/30 + 300. Then R(x) = p * x = x(-x/30 + 300) = -x²/30 + 300x.
The cost function C(x) is typically provided directly.
Profit is always P(x) = R(x) - C(x). Expand and simplify.
The marginal function is the derivative of the original function.
Marginal cost: C'(x)
Marginal revenue: R'(x)
Marginal profit: P'(x)
To estimate the cost, revenue or profit of producing/selling the nth unit, evaluate the marginal function at n - 1.
Example: to estimate the profit on the 20th barrel, compute P'(19).
This gives an approximation. The exact value is P(20) - P(19).
The marginal value is an estimate; the actual change is f(n) - f(n - 1). The two will be close but not identical.
Actual change from the (n-1)th to the nth unit: f(n) - f(n - 1). This is the precise value.
Marginal estimate of the nth unit: f'(n - 1). This is the derivative-based approximation.
Exam problems often ask for both so you can compare them.
Average cost: C-bar(x) = C(x) / x
Average profit: P-bar(x) = P(x) / x
These tell you the per-unit cost or profit at a given production level.
Take the derivative of the average function using the quotient rule (or simplify first, then differentiate).
C-bar'(x) tells you whether per-unit cost is increasing or decreasing.
P-bar'(x) tells you whether per-unit profit is increasing or decreasing.
Interpreting the result: if P-bar'(50) = 0.27, it means that at 50 units, the average profit per unit is increasing by approximately $0.27 for each additional unit produced.
R(x) = p \cdot xP(x) = R(x) - C(x)\bar{C}(x) = \frac{C(x)}{x}\bar{P}(x) = \frac{P(x)}{x}Marginal functions are simply the derivatives of the above:
C'(x),\quad R'(x),\quad P'(x),\quad \bar{C}'(x),\quad \bar{P}'(x)To estimate the cost/revenue/profit of the nth unit, evaluate the marginal function at (n - 1):
\text{Estimated cost of } n\text{th unit} \approx C'(n-1)Marginal analysis is the backbone of production decisions in any business. A factory manager deciding whether to run one more batch, or a retailer deciding whether to stock one more case, is asking exactly the question that P'(x) answers: "Will producing one more unit add to my profit or eat into it?" The sign and size of the marginal profit tell you whether to scale up, hold steady or cut back.
"Profit can't be negative." It absolutely can. A negative profit means the company is losing money at that production level. Revenue and cost cannot be negative, but their difference can be.
Confusing marginal with actual. The marginal cost of the 47th toy is C'(46), not C'(47). You evaluate the derivative at one unit before the one you are estimating. The actual cost of the 47th toy is C(47) - C(46).
Forgetting that marginal is an estimate. Students sometimes present the marginal value as exact. It is an approximation. The exam may ask you to find both the actual and the marginal, then compare.
Mixing up average and marginal. Average cost is the total divided by x. Marginal cost is the derivative. They answer different questions: "How much per unit so far?" versus "How much for one more?"
Building revenue incorrectly from a price-demand equation. If x = 9000 - 30p, do not multiply the entire equation by p. Isolate p first, then multiply by x.
⚠️ You will be asked to build P(x) from a word problem and then find P'(x). Make sure you can do the algebra cleanly before differentiating.
⚠️ Expect a question that asks for both the actual profit/cost of the nth unit AND the marginal estimate, then asks you to compare or interpret.
⚠️ "Does the result make sense?" is a common follow-up when profit is negative. The expected answer: yes, a negative profit means costs exceed revenue, and the company needs to sell more units to break even.
⚠️ Average cost and marginal average cost questions (problems 6 and 7 on the review) are fair game. Know how to form C-bar(x) = C(x)/x and differentiate it.
⚠️ Price-demand problems require you to isolate p before building R(x). Show that algebra step clearly.
True or False: Marginal profit P'(x) gives the exact profit from selling the (x+1)th unit. (False, it is an estimate.)
Fill in the blank: To estimate the cost of the 47th unit, evaluate C'(___). (46)
True or False: If P(50) = -200, the company is losing money at 50 units. (True.)
Fill in the blank: Average cost is calculated as C-bar(x) = ___. (C(x) / x)
True or False: Revenue can be negative. (False. Revenue = price times quantity, both non-negative.)
Q: A company has P(x) = 15x - 0.05x² - 600. What is P(47)? Does the result make sense?
A: P(47) = 15(47) - 0.05(47²) - 600 = 705 - 110.45 - 600 = -5.45, so -$5.45. Yes, it makes sense. Profit can be negative; it means costs currently exceed revenue, and the company needs to sell more handles to turn a profit.
Q: Using the same function, what is the actual profit of the 48th unit?
A: P(48) - P(47). P(48) = 15(48) - 0.05(2304) - 600 = 720 - 115.2 - 600 = 4.80. So P(48) - P(47) = 4.80 - (-5.45) = $10.25.
Q: An oil company sells barrels at $100 each with C(x) = 0.005x³ - 2x + 5000. Find P(x) and P'(x).
A: R(x) = 100x. P(x) = 100x - (0.005x³ - 2x + 5000) = 102x - 0.005x³ - 5000. P'(x) = 102 - 0.015x².
Q: Use marginal analysis to estimate the profit on the 20th barrel.
A: Evaluate P'(19) = 102 - 0.015(361) = 102 - 5.415 = $96.59 (approximately).
Q: Given C(x) = x² - 5x + 1000, find the average cost function and its derivative.
A: C-bar(x) = C(x)/x = x - 5 + 1000/x. C-bar'(x) = 1 - 1000/x².
Q: If P(x) = 30x - 0.03x² - 750, find the average profit at x = 50 and interpret P-bar'(50).
A: P-bar(50) = P(50)/50 = (1500 - 75 - 750)/50 = 675/50 = $13.50 per mower. P-bar'(x) = -0.03 + 750/x². P-bar'(50) = -0.03 + 750/2500 = -0.03 + 0.30 = $0.27. At 50 units, the average profit per mower is increasing by about 27 cents for each additional mower produced.
This material connects directly to differentiation rules (Sections 3.1-3.3), since building marginal functions requires you to take derivatives of polynomials, quotients and products. It also sets up the optimisation problems you will encounter later in the course, where you set marginal profit equal to zero to find the production level that maximises profit.
The average-function work here uses the quotient rule, which reappears heavily in the derivative problems later on the exam.
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