Difficulty: Intermediate-Advanced | Prerequisites: Differentiation rules, curve analysis
This is where all the differentiation rules and curve analysis meet the reason you are taking this course. Business calculus is about turning real questions (how many units should we produce? what price should we charge? how fast is our investment growing?) into functions, differentiating them, and interpreting the results. You will also work with exponential models for compound interest and radioactive decay. If the earlier notes are the toolkit, this is the workbench.
Marginal cost, revenue, and profit are the derivatives of their respective functions. Compound interest uses A = P(1 + r/n)^(nt) for periodic compounding and A = Pe^(rt) for continuous. Optimisation problems ask you to build a function from constraints, differentiate, set equal to zero, and check that you have a max or min.
Marginal cost, C'(x)
The derivative of the cost function. It approximates the cost of producing one additional unit.
Think of it as: how much more it costs to make the next item.
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 P(x) = R(x) − C(x). Equivalently, P'(x) = R'(x) − C'(x).
In simple terms, it tells you how much extra profit you earn from making and selling one more unit.
Average cost
C̄(x) = C(x)/x. The total cost divided by the number of units produced.
Marginal average cost, C̄'(x)
The derivative of the average cost function. Use the quotient rule on C(x)/x.
Compound interest (periodic)
A = P(1 + r/n)^(nt), where P is the principal, r is the annual rate, n is the number of compounding periods per year, and t is time in years.
Continuous compounding
A = Pe^(rt). The limiting case as the number of compounding periods grows without bound.
Think of it as: what happens when interest compounds every instant, not just monthly or daily.
Exponential decay
A = A₀·e^(rt) where r < 0. Used for radioactive decay, depreciation, and similar processes.
Optimisation
The process of finding the input value that maximises or minimises a function, subject to constraints. Build the objective function, differentiate, set the derivative to zero, and verify.
Given C(x) = 4x² + 2x + 100 and price per unit p(x) = 300 − 2x:
Marginal cost: C'(x) = 8x + 2.
Revenue: R(x) = x·p(x) = 300x − 2x². Marginal revenue: R'(x) = 300 − 4x.
Profit: P(x) = R(x) − C(x) = 300x − 2x² − 4x² − 2x − 100 = −6x² + 298x − 100.
Marginal profit: P'(x) = −12x + 298. (Equivalently, R'(x) − C'(x) = (300 − 4x) − (8x + 2) = 298 − 12x.)
Average cost: C̄(x) = C(x)/x = 4x + 2 + 100/x.
Marginal average cost: C̄'(x) = 4 − 100/x² = 4 − 100x⁻².
To estimate the revenue from selling the 61st unit, evaluate R'(60).
For R(x) = 2√x + 7200x, R'(x) = 1/√x + 7200.
R'(60) = 1/√60 + 7200 ≈ 0.129 + 7200 = 7200.13. So the estimated revenue from the 61st unit is approximately $7,200.13.
A = P(1 + r/n)^(nt). With P = 5000, r = 0.06, n = 12, t = 10:
A = 5000(1 + 0.005)^(120) = 5000(1.005)^120 ≈ $9,096.98.
For A = Pe^(rt), doubling means 2P = Pe^(rt), so 2 = e^(rt), giving t = ln(2)/r.
With r = 0.06: t = ln(2)/0.06 ≈ 0.6931/0.06 ≈ 11.55 years.
Given 700 grams decaying to 200 grams in 50 years: 200 = 700·e^(50r).
e^(50r) = 200/700 = 2/7. So 50r = ln(2/7), giving r = ln(2/7)/50 ≈ −1.2528/50 ≈ −0.02506.
Two positive numbers with difference 71, minimise their product. Let the numbers be x and x − 71 (with x > 71).
Product: P(x) = x(x − 71) = x² − 71x. P'(x) = 2x − 71 = 0 gives x = 35.5.
The other number is 35.5 − 71 = −35.5. But the problem says positive numbers, which is contradicted. The answer choices give 35.5 and −35.5, noting that a minimum product exists but one number is negative.
Area = 693 m², three sides at $5/m and one side at $15/m. Let y = expensive side, x = perpendicular side. xy = 693, so x = 693/y.
Cost = 5(2x + y) + 15y = 10x + 20y = 10(693/y) + 20y = 6930/y + 20y.
C'(y) = −6930/y² + 20 = 0. y² = 6930/20 = 346.5. y = √346.5 ≈ 18.6. But checking the answer choices: let y = 21, then x = 693/21 = 33. Cost = 10(33) + 20(21) = 330 + 420 = 750. This is the minimum from the choices offered.
200 rooms at $120/night, all filled. Each $5 increase loses 4 rooms. Service cost $10/room.
Let n = number of $5 increases. Rooms rented: 200 − 4n. Price: 120 + 5n.
Revenue: (200 − 4n)(120 + 5n). Cost: 10(200 − 4n).
Profit: P(n) = (200 − 4n)(120 + 5n) − 10(200 − 4n) = (200 − 4n)(110 + 5n).
Expand: P(n) = 22000 + 1000n − 440n − 20n² = 22000 + 560n − 20n².
P'(n) = 560 − 40n = 0, so n = 14. Price = 120 + 5(14) = $190.
Marginal cost: C'(x) = d/dx[C(x)]
Revenue from price function: R(x) = x · p(x)
Profit: P(x) = R(x) − C(x)
Average cost: C̄(x) = C(x)/x
Marginal estimation: the cost/revenue of the (n+1)th unit ≈ C'(n) or R'(n)
Compound interest (periodic): A = P(1 + r/n)^(nt)
Continuous compounding: A = Pe^(rt)
Doubling time (continuous): t = ln(2)/r
Exponential decay: A = A₀·e^(rt), solve for r using ln
Students sometimes think marginal cost equals the total cost of producing n + 1 units. It does not. Marginal cost approximates the additional cost of one more unit beyond n.
When computing compound interest, students mix up n (compounding periods per year) and t (years). Monthly compounding means n = 12, not n = 120 for 10 years.
In optimisation word problems, students often write the correct objective function but forget to express it in terms of a single variable using the constraint. You cannot differentiate a two-variable expression.
For exponential decay, the rate r is negative. Students frequently drop the negative sign and then wonder why their answer does not match.
⚠️ Marginal analysis is the most business-specific topic on the exam. Know how to interpret C'(x), R'(x), and P'(x) in words.
⚠️ Compound interest and continuous compounding problems are formulaic. Memorise both formulas and practise substituting.
⚠️ Doubling time with continuous compounding: t = ln(2)/r. This comes up frequently.
⚠️ Word problems in optimisation require setting up the function from the problem description. The calculus is usually straightforward; the setup is where marks are lost.
⚠️ Hotel/fencing style problems follow a pattern: define a variable for the quantity you control, express revenue/cost in that variable, differentiate, set to zero.
Fill in the blank: If C(x) = 4x² + 2x + 100, then C'(x) = ___.
True or False: The doubling time for continuous compounding at 5% is ln(2)/5.
Fill in the blank: To estimate the cost of the 101st unit, evaluate ___.
True or False: In A = Pe^(rt), a negative r represents growth.
Q: Given C(x) = 4x² + 2x + 100, find the marginal cost function.
A: C'(x) = 8x + 2.
Q: Find the marginal average cost C̄'(x) for C(x) = 4x² + 2x + 100.
A: C̄(x) = 4x + 2 + 100/x. C̄'(x) = 4 − 100/x² = 4 − 100x⁻².
Q: $5,000 is invested at 6% compounded monthly for 10 years. What is the balance?
A: A = 5000(1.005)^120 ≈ $9,096.98.
Q: At 6% continuous compounding, how long to double?
A: t = ln(2)/0.06 ≈ 11.55 years.
Q: 700 grams decays to 200 grams in 50 years. Find the continuous decay rate.
A: r = ln(200/700)/50 = ln(2/7)/50 ≈ −0.02506.
Q: R(x) = 2√x + 7200x. Estimate the revenue from the 61st unit.
A: R'(x) = 1/√x + 7200. R'(60) ≈ 7200.13, so approximately $7,200.13.
Q: Given p(x) = 300 − 2x and C(x) = 4x² + 2x + 100, find the marginal profit.
A: R(x) = 300x − 2x². P(x) = −6x² + 298x − 100. P'(x) = −12x + 298. (Or equivalently, 298 − 12x.)
Q: A hotel has 200 rooms at $120/night. Each $5 increase loses 4 rooms. Service costs $10/room. What price maximises profit?
A: Let n = number of $5 increases. Profit = (200 − 4n)(110 + 5n). P'(n) = 560 − 40n = 0, so n = 14. Price = $190.
Marginal analysis is the business application of derivatives. Optimisation problems combine differentiation with the second derivative test or closed-interval method from curve analysis. Exponential growth and decay models reuse the chain rule (differentiating e^(rt)) and logarithmic properties.
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