Difficulty: Intermediate | Prerequisites: Basic differentiation (power rule), understanding of cost functions
Marginal analysis uses derivatives to estimate how costs, revenues and profits change when you produce one more unit. Instead of recalculating the entire cost function, you evaluate its derivative at the current output level to get a quick approximation. The trade-off: the marginal (derivative) answer is an estimate, not the exact figure, but it is far faster to compute and close enough for most business decisions.
Marginal revenue
The additional revenue gained by producing and selling one more unit of a good or service. Formally, it is the derivative of the revenue function, R'(x). Think of it as: "How much extra money comes in if we sell one more?"
Marginal profit
The increase in profit resulting from producing one additional unit. Formally, P'(x) = R'(x) - C'(x). In simple terms, this means: the extra profit you pocket from making one more item, after subtracting the extra cost.
Marginal cost
The derivative of the total cost function, C'(x). It represents the cost added by producing one additional unit of a good or service. Think of it as: "How much more does it cost to make the next one?"
Average cost (per unit)
Total cost divided by the number of units produced: C̄(x) = C(x) / x. In simple terms, this means: if you spread all your costs evenly across every unit, how much does each one cost?
Marginal average cost
The derivative of the average cost function, C̄'(x). It tells you how the average cost per unit changes as you produce one more item. Think of it as: "Does making one more unit push my per-unit cost up or down?"
Exact cost of the nth unit
The actual cost of producing unit number n, calculated as C(n) - C(n - 1). This gives the precise figure, not an approximation.
Exact cost of the nth unit is computed by evaluating C(n) - C(n - 1). This requires plugging in twice and subtracting.
Approximate cost of the nth unit uses the marginal cost function: just evaluate C'(n). One plug-in, done.
The marginal approach is an approximation because C'(x) gives the instantaneous rate of change, not the actual discrete jump from unit n - 1 to unit n.
For most business cost functions, the two values are very close. The difference shrinks when the cost curve is nearly linear in that region.
Given: C(x) = 1000 + 100x - 0.25x²
Finding the exact cost of the 51st guitar:
C(51) = 1000 + 100(51) - 0.25(51)² = 5449.75
C(50) = 1000 + 100(50) - 0.25(50)² = 5375
Exact cost of the 51st guitar = 5449.75 - 5375 = 74.75
Finding the marginal cost function:
Differentiate C(x): C'(x) = 100 - 0.5x
This is the marginal cost function, valid for any output level x.
Using marginal analysis to approximate the 51st guitar:
C'(51) = 100 - 0.5(51) = 74.5
Compare: exact cost was 74.75, marginal estimate is 74.5. The difference is only 0.25.
Average cost = C(x) / x. It tells you the cost per unit if costs are spread evenly.
Marginal average cost = derivative of the average cost function. It tells you how the per-unit cost is shifting as output changes.
Given: C(x) = 10,000 + 20x
Average cost at 1,000 units:
C(1000) = 10,000 + 20(1000) = 30,000
Average cost = 30,000 / 1000 = 30 per board game
Note: the student's work in the source showed C(1000) = 20,010, which appears to have used C(x) = 10,000 + 20x but divided incorrectly or used a slightly different function. The key method is: compute C(x), then divide by x.
Marginal average cost at 1,000 units:
Average cost function: C̄(x) = C(x)/x = 10,000/x + 20
Differentiate: C̄'(x) = -10,000/x²
At x = 1000: C̄'(1000) = -10,000/1,000,000 = -0.01
Interpretation: at a production level of 1,000 units, each additional unit produced reduces the average cost by approximately £0.01. The average cost is falling, though very slowly.
Estimating average cost at 1,001 units:
C̄(1001) ≈ C̄(1000) + C̄'(1000) = 30 + (-0.01) = 29.99
The average cost per board game drops fractionally when you produce one more unit. This is the effect of spreading the fixed cost (10,000) across more units.
\text{Exact cost of unit } n = C(n) - C(n-1)\text{Marginal cost} = C'(x) = \frac{dC}{dx}\text{Average cost} = \bar{C}(x) = \frac{C(x)}{x}\text{Marginal average cost} = \bar{C}'(x) = \frac{d}{dx}\left(\frac{C(x)}{x}\right)\text{Marginal revenue} = R'(x) = \frac{dR}{dx}\text{Marginal profit} = P'(x) = R'(x) - C'(x)Confusing marginal cost with total cost. Students sometimes evaluate C(n) when asked for the marginal cost of the nth unit. Marginal cost is C'(n), the derivative evaluated at n, not the total cost at n units.
Evaluating C'(n) at n - 1 instead of n. The convention in this course is to evaluate the marginal cost at the unit number itself: C'(51) for the 51st unit, not C'(50). Some textbooks differ, so check your professor's convention.
Thinking marginal cost gives the exact cost. The marginal cost is an approximation of the cost of the next unit. The exact cost is always C(n) - C(n - 1). The two are close but not identical.
Forgetting to divide by x for average cost. Average cost is C(x)/x, not just C(x). Students sometimes report the total cost when asked for the average.
Treating marginal average cost as the same as marginal cost. These are different derivatives. Marginal cost = C'(x). Marginal average cost = d/dx[C(x)/x]. They answer different questions.
⚠️ Expect a question that asks for both the exact cost and the marginal approximation of the nth unit, then asks you to compare them. Know how to do both and explain why they differ.
⚠️ Average cost questions often appear with linear cost functions (like C(x) = fixed + variable·x) because the algebra stays clean. Be comfortable with the quotient rule or rewriting C(x)/x before differentiating.
⚠️ Interpretation questions are common: "What does C'(200) = 45 mean?" You need to write a sentence: "Producing the 200th unit adds approximately $45 to total cost."
⚠️ When marginal average cost equals zero, average cost is at its minimum. This is a classic optimisation question that builds directly on this material.
True or false: Marginal cost gives the exact cost of producing the next unit. (False, it is an approximation.)
Fill in the blank: The exact cost of the nth unit is calculated as ____. (C(n) - C(n - 1))
True or false: If C(x) = 5000 + 30x, the marginal cost function is C'(x) = 30. (True.)
Fill in the blank: Average cost equals ____ divided by ____. (C(x) divided by x)
True or false: When marginal average cost is negative, average cost is decreasing. (True.)
Q: Given C(x) = 800 + 50x - 0.1x², find the exact cost of producing the 101st unit.
A: C(101) - C(100) = [800 + 50(101) - 0.1(101)²] - [800 + 50(100) - 0.1(100)²] = [800 + 5050 - 1020.1] - [800 + 5000 - 1000] = 4829.9 - 4800 = 29.9.
Q: For the same cost function, find the marginal cost at x = 101 and compare it to the exact cost.
A: C'(x) = 50 - 0.2x. C'(101) = 50 - 0.2(101) = 50 - 20.2 = 29.8. The marginal estimate (29.8) is close to the exact cost (29.9), differing by 0.1.
Q: A company's total cost function is C(x) = 20,000 + 15x. What is the average cost per unit at x = 500? At x = 2000? What happens to average cost as production increases?
A: At x = 500: C̄ = (20,000 + 7,500)/500 = 55. At x = 2000: C̄ = (20,000 + 30,000)/2000 = 25. Average cost falls as production rises because the fixed cost (20,000) is spread over more units.
Q: Explain in one sentence what C'(300) = 42 means in a business context.
A: Producing the 300th unit adds approximately $42 to the company's total cost.
Q: If the average cost function is C̄(x) = 10,000/x + 20, find the marginal average cost and interpret C̄'(1000).
A: C̄'(x) = -10,000/x². At x = 1000: C̄'(1000) = -10,000/1,000,000 = -0.01. This means that at a production level of 1,000 units, producing one more unit reduces the average cost by approximately $0.01.
This material connects directly to profit maximisation: a firm maximises profit where marginal revenue equals marginal cost, R'(x) = C'(x). That result is the centrepiece of business calculus optimisation problems.
Average cost analysis ties into economies of scale. When fixed costs are high relative to variable costs, average cost drops steeply as output rises, which is the mathematical basis for why larger firms can price more competitively.
The derivative interpretation here (rate of change at a point as an approximation of the discrete change) is the same concept used in related rates, elasticity of demand, and sensitivity analysis throughout the course.
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