Difficulty: Intermediate | Prerequisites: Basic differentiation (power rule, polynomial derivatives), algebra of polynomial functions
Marginal analysis is where calculus meets business decision-making. In MAC 2233, you learn to model a company's costs, revenue and profit as functions of the number of units produced or sold, then use derivatives to estimate what happens at the margin, that is, when one more unit is made or sold. This material sits at the intersection of Sections 2.7 and 3.1, bridging pure differentiation technique with applied business problems. You should already be comfortable taking derivatives of polynomials using the power rule before tackling these problems.
Profit = Revenue minus Cost. The marginal version of any business function is just its derivative, and it estimates the change from producing or selling one additional unit. Plug in the current quantity to get the marginal value, then use that to approximate the next unit's impact without recalculating the entire function.
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.
In simple terms, this is the money coming in before you subtract any costs.
Cost function, C(x)
The total cost of producing x units. Typically given as a polynomial in x.
Think of it as everything the company spends to make those units: materials, labour, overhead.
Profit function, P(x)
The difference between revenue and cost: P(x) = R(x) − C(x).
In simple terms, this is what the company keeps after paying all production costs.
Marginal cost, C'(x)
The derivative of the cost function. It approximates the cost of producing the (x + 1)th unit.
Think of it as: "How much more does it cost to make one more?"
Marginal revenue, R'(x)
The derivative of the revenue function. It approximates the additional revenue from selling one more unit.
Marginal profit, P'(x)
The derivative of the profit function. It estimates the change in profit from producing and selling one additional unit.
Average cost, C̄(x)
Total cost divided by the number of units: C̄(x) = C(x) / x.
In simple terms, this is the per-unit cost when you spread total spending evenly across all units produced.
Average profit, P̄(x)
Total profit divided by the number of units: P̄(x) = P(x) / x.
Marginal average cost, C̄'(x)
The derivative of the average cost function. It estimates how the per-unit cost changes when one more unit is produced.
Revenue is straightforward when price per unit is fixed: R(x) = (price per unit) × x
Example: an oil company sells barrels at $100 each, so R(x) = 100x
Cost is typically given in the problem as a polynomial in x
Example: C(x) = 0.005x³ − 2x + 5000
Profit is always Revenue minus Cost: P(x) = R(x) − C(x)
Example: P(x) = 100x − (0.005x³ − 2x + 5000) = −0.005x³ + 102x − 5000
Careful with signs when distributing the negative across all terms of C(x)
The marginal cost is C'(x), found by differentiating C(x) using the power rule
The marginal revenue is R'(x)
The marginal profit is P'(x) = R'(x) − C'(x)
These derivatives tell you the instantaneous rate of change, which for business purposes approximates the effect of one additional unit
To estimate the profit (or cost, or revenue) of the (x + 1)th unit, evaluate the marginal function at x
Example: to estimate the profit of the 21st barrel, compute P'(20), not P'(21)
This is because the derivative at x = 20 approximates the change from unit 20 to unit 21
To find the exact value of the (x + 1)th unit, compute f(x + 1) − f(x)
Example: exact profit of the 48th door handle = P(48) − P(47)
The marginal estimate and the exact value will be close but not identical
Average cost: C̄(x) = C(x) / x
Example: C(x) = x² − 5x + 1000, so C̄(x) = x − 5 + 1000/x
Average profit: P̄(x) = P(x) / x
Marginal average cost: take the derivative of C̄(x), not of C(x)
Example: if C̄(x) = x − 5 + 1000/x, then C̄'(x) = 1 − 1000/x²
Sometimes you are given a price-demand equation like x = 9000 − 30p
Solve for p in terms of x to get the price function: p = (9000 − x) / 30 = 300 − x/30
Revenue then becomes R(x) = x · p(x) = 300x − x²/30
The domain of p(x) is determined by requiring p ≥ 0 and x ≥ 0
Profit is still R(x) − C(x), and marginal profit is P'(x)
P(x) = R(x) - C(x)\bar{C}(x) = \frac{C(x)}{x}\bar{P}(x) = \frac{P(x)}{x}\text{Marginal cost} = C'(x) \approx C(x+1) - C(x)\text{Marginal profit} = P'(x) \approx P(x+1) - P(x)R(x) = x \cdot p(x) \quad \text{(when price depends on demand)}Key point: to estimate the value of the (n)th unit, evaluate the marginal function at (n − 1).
Manufacturers use marginal cost analysis to decide whether ramping up production is worth it: if the marginal cost of the next unit exceeds the marginal revenue, the company loses money by making it. Airlines, for instance, use this logic to price last-minute seats, selling them cheaply as long as the fare exceeds the marginal cost of carrying one more passenger.
Students often evaluate the marginal function at x = n to estimate the nth unit. They should evaluate at x = n − 1. To estimate the cost of the 47th toy, compute C'(46), not C'(47).
Students frequently forget to distribute the negative sign when computing P(x) = R(x) − C(x). Every term inside C(x) changes sign, not just the first one.
Average cost is not the same as marginal cost. C̄(x) = C(x)/x tells you the per-unit cost spread across all units; C'(x) tells you the approximate cost of making one more.
A negative profit does not mean the formula is wrong. It means the company is operating at a loss at that production level. Always check whether the result "makes sense" by considering whether costs exceed revenue.
⚠️ "Use marginal analysis to estimate..." is exam language for: take the derivative, plug in the current quantity (one less than the unit you are estimating), and interpret the result as an approximation.
⚠️ You will almost certainly be asked to find P(x), then P'(x), then use P'(x) to estimate a specific unit. Know the full chain.
⚠️ "Does the result make sense?" is a prompt to check the sign. Negative profit = loss. Negative marginal profit = profit is decreasing.
⚠️ Price-demand problems combine algebra (solving for p) with calculus (differentiating R(x) = x · p(x)). Expect at least one on the exam.
True or False: To estimate the cost of the 51st unit, you evaluate C'(51). (False, you evaluate C'(50).)
Fill in the blank: P(x) = ______ − C(x). (R(x))
True or False: Marginal profit equals the derivative of profit. (True.)
Fill in the blank: Average cost is calculated as C̄(x) = C(x) / ______. (x)
True or False: If P(47) = −5.45, the company is making a profit at 47 units. (False, it is operating at a loss.)
Q: A company has P(x) = 15x − 0.05x² − 600. What is the profit when 47 handles are sold? Does the result make sense?
A: P(47) = 15(47) − 0.05(47²) − 600 = 705 − 110.45 − 600 = −5.45. The company is at a small loss, which makes sense if production is not yet at a profitable scale.
Q: Given C(x) = 0.1x³ − 2x² + 10000, what is the actual cost of producing the 47th toy?
A: C(47) − C(46) = [0.1(47³) − 2(47²) + 10000] − [0.1(46³) − 2(46²) + 10000]. Work through the arithmetic: C(47) = 10383.3 − 4418 + 10000 = 15965.3, C(46) = 9733.6 − 4232 + 10000 = 15501.6. So the actual cost of the 47th toy is approximately $463.70.
Q: A company's revenue function is R(x) = 2x² + x and its cost function is C(x) = 60x + 72000. Find the profit function and the marginal profit function.
A: P(x) = R(x) − C(x) = 2x² + x − 60x − 72000 = 2x² − 59x − 72000. The marginal profit is P'(x) = 4x − 59.
Q: If x = 9000 − 30p, express p as a function of x and state the domain.
A: Solving for p gives p = (9000 − x)/30 = 300 − x/30. The domain requires p ≥ 0 and x ≥ 0, so 0 ≤ x ≤ 9000.
Q: Given P(x) = 30x − 0.03x² − 750, find the average profit per mower when 50 mowers are produced.
A: P̄(x) = P(x)/x. First, P(50) = 1500 − 75 − 750 = 675. Then P̄(50) = 675/50 = $13.50 per mower.
Marginal analysis relies on differentiation rules covered in Sections 3.1 through 3.3, so if derivative techniques feel shaky, revise those first. The concept of marginal analysis reappears in optimisation problems (finding maximum profit, minimum cost) later in the course, where you set the marginal function to zero and solve. Compound interest and exponential growth models (Section 3.2) use similar "rate of change" reasoning, applied to financial functions instead of polynomial business functions.
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