Difficulty: Intermediate to Advanced | Prerequisites: Critical points and extrema notes, derivative rules notes
Optimisation problems are where calculus meets the real world in MAT 210. The question is always some version of "what input gives the best output?" and the method is always the same: build a function, take its derivative, set it to zero, and check your answer. Revenue maximisation is the most common application. Rate-of-change problems using the product rule are the other major type. Both require comfort with derivative rules and critical-point analysis.
Revenue = price times quantity. To maximise revenue, write R(p) = p · q(p), differentiate, set R'(p) = 0, and solve for p. For product-rule rate-of-change problems, use v'(t) = n'(t) · c(t) + n(t) · c'(t) and evaluate at the given time. Always include units in your final answer.
Demand function, q(p)
The number of units sold as a function of price. As price goes up, demand typically goes down. Can be linear (e.g. q(p) = 150000 − 2500p) or nonlinear (e.g. q(p) = 750 − 10p²).
Revenue function, R(p)
Total revenue from sales: R(p) = p · q(p), where p is the price per unit and q(p) is the quantity sold. Think of it as: revenue is what you charge times how many you sell.
Rate of change
The derivative of a quantity with respect to time (or another variable). Tells you how fast something is increasing or decreasing at a specific moment. Units matter: dollars per day, litres per hour, and so on.
Product rule for rates
When a total quantity is the product of two time-dependent factors, v(t) = n(t) · c(t), the rate of change is v'(t) = n'(t) · c(t) + n(t) · c'(t). In simple terms, both factors are changing, so you account for each one's contribution.
Marginal revenue
The derivative of the revenue function. It tells you how much additional revenue you gain (or lose) from a small increase in price or quantity.
Worked example: q(p) = 750 − 10p², R(p) = p · q(p) = 750p − 10p³.
R'(p) = 750 − 30p²
Set R'(p) = 0: 750 = 30p², so p² = 25, giving p = 5 (price must be positive)
Check: R''(p) = −60p, so R''(5) = −300 < 0, confirming a maximum
The price that maximises revenue is p = 5 dollars
Worked example: q(p) = 150000 − 2500p, R(p) = p · q(p) = 150000p − 2500p².
R'(p) = 150000 − 5000p
Set R'(p) = 0: 150000 = 5000p, so p = 30 dollars
Maximum revenue: R(30) = 150000(30) − 2500(900) = 4,500,000 − 2,250,000 = $2,250,000
Interpretation: at a price of $30 per unit, the company earns the maximum possible revenue of $2,250,000
Worked example: n(t) = 16t² + 2t hats, c(t) = ln(230t) dollars per hat, v(t) = n(t) · c(t) total cost.
Find v'(16):
n'(t) = 32t + 2
c'(t) = 1/t (since the derivative of ln(230t) = 230/(230t) = 1/t)
v'(t) = n'(t) · c(t) + n(t) · c'(t) = (32t + 2) · ln(230t) + (16t² + 2t) · (1/t)
At t = 16:
n'(16) = 32(16) + 2 = 514
c(16) = ln(3680) ≈ 8.2106
n(16) = 16(256) + 32 = 4128
c'(16) = 1/16 = 0.0625
v'(16) = 514 · 8.2106 + 4128 · 0.0625 = 4220.25 + 258.00 = 4478.25
Rounding: the total cost is increasing at approximately 4478.28 dollars per day on the 16th day. (The small rounding difference comes from the precise value of ln(3680).)
Every rate-of-change answer needs three things: the numerical value, the units, and whether the quantity is increasing or decreasing. If v'(t) > 0, the quantity is increasing. If v'(t) < 0, it is decreasing. Units follow from the original functions: if v is in dollars and t is in days, then v'(t) is in dollars per day.
Revenue optimisation is how companies set prices for everything from earbuds to airline tickets. The maths is exactly what you see on this exam: model demand, build a revenue function, and find the peak. Rate-of-change problems model how costs, volumes, or populations change in real time, which matters in manufacturing, environmental science, and finance.
Students often forget to write R(p) = p · q(p) as a single expression in p before differentiating. You cannot differentiate R(p) = p · q(p) as if p and q are independent; q is a function of p.
Students sometimes give the critical value (e.g. p = 30) as the final answer when the question asks for the maximum revenue. The price is one answer; the revenue R(30) is another. Read the question.
In product-rule rate problems, students sometimes forget to include both terms. The total rate of change has two pieces, n'(t) · c(t) and n(t) · c'(t), and both matter.
Students often drop units from rate-of-change answers. "4478.28" is incomplete. "4478.28 dollars per day" is the answer.
⚠️ Revenue maximisation appears in both multiple-choice and free-response form. Know the steps cold: write R(p), differentiate, set to zero, solve, and verify.
⚠️ Product-rule rate problems (like the hat manufacturer question) require careful arithmetic. Show every step so partial credit is possible even if you make a calculation error.
⚠️ Always state whether a quantity is increasing or decreasing, and always include units.
Fill in the blank: Revenue = ______ times ______. Answer: price times quantity (R = p · q).
True or False: If R'(p) = 0 and R''(p) < 0, then p gives a maximum revenue. Answer: True.
True or False: The derivative of ln(230t) with respect to t is 230/t. Answer: False. It is 1/t. (The 230 cancels: 230/(230t) = 1/t.)
Fill in the blank: If v(t) = n(t) · c(t), then v'(t) = ______ + ______. Answer: n'(t) · c(t) + n(t) · c'(t).
Q: A company's demand function is q(p) = 750 − 10p². Find the price that maximises revenue.
A: R(p) = 750p − 10p³. R'(p) = 750 − 30p² = 0, so p² = 25, giving p = 5 dollars.
Q: An audio company has demand q(p) = 150000 − 2500p. Find the price that maximises revenue and state the maximum revenue.
A: R(p) = 150000p − 2500p². R'(p) = 150000 − 5000p = 0, giving p = $30. Maximum revenue: R(30) = $2,250,000. At a price of $30, the company earns the highest possible revenue.
Q: Given n(t) = 16t² + 2t and c(t) = ln(230t), with v(t) = n(t) · c(t), find v'(16) and interpret the result.
A: v'(16) ≈ 4478.28. The total cost is increasing at a rate of approximately $4478.28 per day on the 16th day.
Q: V(t) = t³e^(0.02t²) gives the volume of water in litres, with t in hours. Find the rate of change at t = 5 hours.
A: V'(t) = e^(0.02t²)(3t² + 0.04t⁴). V'(5) = e^(0.5)(75 + 25) = 100e^(0.5) ≈ 164.87 litres per hour. The volume is increasing at about 164.87 litres per hour at t = 5.
Optimisation problems are critical points in context. Every revenue-max or cost-min problem uses the same machinery from the critical points notes: set the derivative to zero, solve, classify. The product rule rate-of-change problems connect directly to the derivative rules notes, layering the product rule with chain rule or logarithmic derivatives. Later in the course, second derivatives provide another classification tool.
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