Difficulty: Intermediate | Prerequisites: All derivative rules, algebra (factoring quadratics, solving equations).
This is where the derivative stops being an abstract operation and starts doing something useful. Every optimisation problem on the exam follows the same skeleton: build a function, differentiate, set the derivative to zero, solve, and verify. Tangent-line problems test whether you can connect the derivative's value at a point to the equation of a line. Marginal analysis links calculus to economics. If the derivative rules feel shaky, solidify those before tackling this material.
TL;DR: Applications of derivatives cover tangent lines, the second derivative test for classifying extrema, optimisation word problems (minimum cost, maximum profit), and marginal revenue. The workflow is always the same: differentiate, set equal to zero, solve, interpret.
Critical number
A value x = c in the domain of f where f'(c) = 0 or f'(c) is undefined. Think of it as a candidate for a local maximum, local minimum, or inflection point.
Second derivative test
A method for classifying critical numbers. If f''(c) > 0, the function is concave up at c, so c is a local minimum. If f''(c) < 0, concave down, so c is a local maximum. If f''(c) = 0, the test is inconclusive.
Tangent line
The line that touches the curve at exactly one point and has a slope equal to the derivative at that point. Its equation in slope-intercept form is y = f'(a)(x − a) + f(a).
Marginal revenue
The derivative of the revenue function, R'(x). It approximates the revenue earned by selling one additional unit. In simple terms, it is the "next-unit revenue estimate."
Optimisation
Finding the input value that makes a function as large or as small as possible, often subject to a constraint. In business calculus, this means minimising cost, maximising revenue, or maximising profit.
Find f'(x) and set it to zero to locate critical numbers.
Compute f''(x) and evaluate it at each critical number.
f''(c) > 0: local minimum at x = c.
f''(c) < 0: local maximum at x = c.
f''(c) = 0: test is inconclusive (use the first derivative test instead).
Example: f''(x) = 12x − 12. Critical numbers at x = 1 and x = 5. At x = 5: f''(5) = 60 − 12 = 48 > 0, so x = 5 is a local minimum.
Step 1: find f(a) to get the point on the curve.
Step 2: find f'(a) to get the slope.
Step 3: write y = f'(a)(x − a) + f(a), then simplify to slope-intercept form.
Example: y = 2x³ − 3x² − 12x + 4 at x = 1.
f(1) = 2 − 3 − 12 + 4 = −9. Point: (1, −9).
f'(x) = 6x² − 6x − 12. f'(1) = 6 − 6 − 12 = −12. Slope: −12.
Tangent line: y = −12(x − 1) − 9 = −12x + 3.
Horizontal tangent lines: set f'(x) = 0. For f'(x) = 6x² − 6x − 12 = 0, divide by 6: x² − x − 2 = 0. Factor: (x − 2)(x + 1) = 0. So x = 2 and x = −1.
"Two numbers whose difference is 71, product is minimum."
Let x − y = 71, so y = x − 71. Product: P = x · y = x(x − 71) = x² − 71x.
P'(x) = 2x − 71 = 0, so x = 35.5 and y = 35.5 − 71 = −35.5.
The numbers are 35.5 and −35.5.
Area = xy = 693. Three sides cost $8/m, one side costs $14/m.
Label the sides: two widths (y) and one length (x) at $8 each, and one length (x) at $14.
Cost: C = 8(2y + x) + 14x = 16y + 8x + 14x = 16y + 22x.
Constraint: y = 693/x. Substitute: C(x) = 16(693/x) + 22x = 11088/x + 22x.
C'(x) = −11088/x² + 22 = 0. Solve: x² = 11088/22 = 504, so x = √504 ≈ 22.45 m.
y = 693/22.45 ≈ 30.87 m.
At $160 per room, all rooms are filled. Each $10 increase loses 4 rooms. Each room costs $40 to service.
Let n = number of $10 increases. Price per room: 160 + 10n. Rooms rented: R₀ − 4n (where R₀ is total rooms at full capacity).
Profit per room: (160 + 10n − 40) = 120 + 10n. Total profit: P(n) = (120 + 10n)(R₀ − 4n).
Since the problem asks for price only: expand, differentiate, set to zero. The exam expects the answer in terms of the charge per room. The structure: P = (price − 40)(rooms), differentiate with respect to price or the number of increases, solve.
With the standard setup, the optimal price depends on R₀. For a typical version, the answer works out to a specific dollar amount once R₀ is known or cancels.
R(x) = 2x² + 7200x. Actual revenue from the 61st unit: R(61) − R(60).
R(61) = 2(3721) + 7200(61) = 7442 + 439200 = 446642.
R(60) = 2(3600) + 7200(60) = 7200 + 432000 = 439200.
Actual = 446642 − 439200 = 7442.
Marginal estimate: R'(x) = 4x + 7200. R'(60) = 240 + 7200 = 7440.
The marginal estimate (7440) is close to the actual (7442). The derivative approximates the next unit's revenue.
y - f(a) = f'(a)(x - a) \quad \text{(tangent line at } x = a\text{)}\text{Second derivative test: } f''(c) > 0 \Rightarrow \text{local min}, \quad f''(c) < 0 \Rightarrow \text{local max}\text{Actual revenue from unit } (n+1) = R(n+1) - R(n)\text{Marginal revenue estimate for unit } (n+1) \approx R'(n)Optimisation is how businesses decide how many units to produce, what price to charge, and how to allocate resources. The fence problem is a stand-in for any constrained-resource question: given a fixed budget or area, what dimensions minimise cost or maximise output? Marginal analysis is the language of micro-economics and is used daily in pricing, production, and inventory decisions.
Students often confuse the tangent line's slope with the function's value. The slope comes from f'(a), the y-value comes from f(a). Both are needed.
In optimisation, students forget to substitute the constraint before differentiating. You must reduce the problem to one variable.
The second derivative test gives no answer when f''(c) = 0. Students sometimes assume this means "no extremum," but it means you need a different method.
For marginal revenue, students confuse R'(60) with R'(61). The convention is that R'(60) estimates the revenue from the 61st unit (the next unit after 60).
⚠️ Tangent-line problems always ask for slope-intercept form (y = mx + b). Do not leave the answer in point-slope form.
⚠️ For the second derivative test, state the conclusion explicitly: "local minimum" or "local maximum." Showing the sign of f''(c) alone is not enough for full marks.
⚠️ Optimisation word problems are typically 7-mark long-answer questions. Show the constraint, the objective function, the derivative, the critical point, and the verification.
⚠️ Marginal analysis questions often ask for both the actual and the estimated value. Compute both and note they are close.
True or False: If f''(c) > 0, then x = c is a local maximum.
False. f''(c) > 0 means concave up, which gives a local minimum.
Fill in the blank: The slope of the tangent line at x = a is given by ___.
f'(a).
True or False: R'(60) estimates the revenue from producing the 60th unit.
False. R'(60) estimates the revenue from the 61st unit.
Q: Given f''(x) = 12x − 12 and a critical number at x = 5, classify the extremum.
A: f''(5) = 60 − 12 = 48 > 0. Since the second derivative is positive, x = 5 is a local minimum.
Q: Find the tangent line to y = 2x³ − 3x² − 12x + 4 at x = 1.
A: f(1) = −9. f'(x) = 6x² − 6x − 12, so f'(1) = −12. Tangent line: y = −12x + 3.
Q: Two numbers have a difference of 71. What values minimise their product?
A: Let x − y = 71. Product P = x(x − 71). P' = 2x − 71 = 0 gives x = 35.5, y = −35.5.
Q: R(x) = 2x² + 7200x. Find the actual and marginal revenue for the 61st unit.
A: Actual: R(61) − R(60) = 446642 − 439200 = 7442. Marginal: R'(60) = 4(60) + 7200 = 7440.
Q: For y = 2x³ − 3x² − 12x + 4, find the x-values where the tangent is horizontal.
A: Set f'(x) = 6x² − 6x − 12 = 0. Divide by 6: x² − x − 2 = (x − 2)(x + 1) = 0. So x = 2 and x = −1.
Optimisation uses derivatives (from the derivative rules topic) and requires continuity (from the limits topic). Marginal analysis connects to integration later: the total revenue over an interval is the integral of marginal revenue. The second derivative test also connects to curve sketching and concavity, which are tested alongside inflection points.
critical number, critical point, second derivative test, local minimum, local maximum, concavity, tangent line, slope-intercept form, horizontal tangent, optimisation, optimization, fence problem, minimum cost, maximum profit, marginal revenue, marginal cost, marginal analysis, MAC 2233, calculus for business, FIU