Difficulty: Intermediate to Advanced | Prerequisites: Derivatives and Differentiation Rules notes
This is where calculus starts earning its keep. Once you can compute derivatives, the next question is: what can you do with them? This section covers three big application areas from the MAT 210 syllabus, all of which appear on both the midterm and the final. Marginal analysis uses derivatives to answer business questions about cost, revenue, and profit. Graph analysis uses the first and second derivatives to map out a function's shape. Optimisation finds the maximum or minimum value of a function, which is the core skill in any problem that asks for "the best" or "the most efficient" outcome.
Derivatives let you answer three kinds of practical question: how much does cost, revenue, or profit change when you produce one more unit (marginal analysis), where does a function reach its highest or lowest value (optimisation), and what does a function's graph look like without plotting dozens of points (graph analysis). Mastering these applications is worth at least half of the MAT 210 final.
Marginal cost, C'(x)
The derivative of the total cost function. It approximates the cost of producing one additional unit when you are already producing x units. Think of it as the price tag on "one more."
Marginal revenue, R'(x)
The derivative of the total revenue function. It approximates the additional revenue earned from selling one more unit.
Marginal profit, P'(x)
The derivative of the profit function P(x) = R(x) - C(x). Positive marginal profit means producing one more unit adds to your profit; negative means it subtracts.
Critical point
A value of x where f'(x) = 0 or f'(x) is undefined. These are the only candidates for local maxima or minima.
Local maximum
A point where f(c) is greater than or equal to the values of f at all nearby points. The function changes from increasing to decreasing at a local max.
Local minimum
A point where f(c) is less than or equal to the values of f at all nearby points. The function changes from decreasing to increasing at a local min.
Absolute (global) extrema
The largest (absolute maximum) and smallest (absolute minimum) values of f on a given interval. On a closed interval, these occur at critical points or at the endpoints.
Concavity
The direction a graph curves. Concave up means the graph bends upward (like a bowl); concave down means it bends downward (like a hill). Determined by the sign of the second derivative.
Inflection point
A point where the concavity changes. At an inflection point, f''(x) = 0 or is undefined, and the sign of f'' switches from one side to the other.
First derivative test
A method for classifying critical points: check the sign of f' to the left and right of the critical point. If f' changes from positive to negative, it is a local max. If negative to positive, a local min.
Second derivative test
A shortcut: at a critical point where f'(c) = 0, if f''(c) > 0 the point is a local min (concave up), and if f''(c) < 0 it is a local max (concave down). If f''(c) = 0, the test is inconclusive.
If C(x) is total cost and R(x) is total revenue for x units, then profit is P(x) = R(x) - C(x).
Marginal cost C'(x) ≈ the cost of the (x + 1)th unit. Marginal revenue R'(x) ≈ the revenue from selling the (x + 1)th unit.
Profit is maximised where marginal revenue equals marginal cost: R'(x) = C'(x). At that point, the gain from one more unit exactly covers its cost.
Example: if C(x) = 0.01x² + 5x + 100 and R(x) = 15x - 0.02x², then P(x) = -0.03x² + 10x - 100. P'(x) = -0.06x + 10 = 0 gives x ≈ 167 units.
Interpreting marginal values: if C'(200) = 12, that means producing the 201st unit costs approximately $12.
Elasticity E = -(p/q) · (dq/dp), where q = f(p) is the demand function.
|E| > 1: elastic demand (a price increase reduces revenue).
|E| < 1: inelastic demand (a price increase raises revenue).
|E| = 1: unit elastic (revenue is at its maximum).
Increasing and decreasing intervals: f is increasing where f'(x) > 0 and decreasing where f'(x) < 0. Find critical points first, then test the sign of f' in each interval between them.
First derivative test: at a critical point c, if f' changes from + to -, c is a local max. If f' changes from - to +, c is a local min. If f' does not change sign, c is neither.
Concavity: f is concave up where f''(x) > 0 (tangent lines lie below the curve) and concave down where f''(x) < 0 (tangent lines lie above the curve).
Inflection points: occur where f'' changes sign. Set f''(x) = 0, solve, and check that the sign of f'' switches.
Second derivative test: at a critical point where f'(c) = 0, if f''(c) > 0 then c is a local min, if f''(c) < 0 then c is a local max. Saves you from doing a sign chart of f' in many cases.
Finding critical points: set f'(x) = 0 and solve. Also note any x where f'(x) is undefined.
Closed interval method (for absolute extrema on [a, b]): evaluate f at each critical point in [a, b] and at the endpoints a and b. The largest value is the absolute max; the smallest is the absolute min.
Applied optimisation: translate the word problem into a function of one variable, find its domain, take its derivative, set it to zero, and verify your answer is a max or min (using the first or second derivative test).
Example: a farmer has 200 metres of fencing and wants to enclose the largest rectangular area against a barn wall (so only three sides need fencing). If the two sides perpendicular to the barn are x metres each, the parallel side is 200 - 2x. Area A(x) = x(200 - 2x) = 200x - 2x². A'(x) = 200 - 4x = 0, so x = 50. A(50) = 5000 m². Since A''(x) = -4 < 0, this is a maximum.
P(x) = R(x) - C(x)\text{Profit maximised when } R'(x) = C'(x)\text{Critical points: solve } f'(x) = 0Second derivative test at a critical point c where f'(c) = 0:
f''(c) > 0 implies local minimum (concave up)
f''(c) < 0 implies local maximum (concave down)
f''(c) = 0 implies test is inconclusive
E = -\frac{p}{q} \cdot \frac{dq}{dp} \quad \text{(elasticity of demand)}Closed interval method on [a, b]: compare f(a), f(b), and f(c) for each critical point c in (a, b).
Marginal analysis is the standard framework in business for deciding whether to produce one more unit of a product. Any firm that can model its cost and revenue functions uses derivatives to find the production level that maximises profit.
Optimisation appears in engineering (minimising material for a given volume), logistics (shortest route problems), and finance (maximising return subject to constraints).
Assuming that a critical point is automatically a maximum or minimum. It might be neither (a saddle point or an inflection point with a horizontal tangent). You must verify with the first or second derivative test.
Confusing local and absolute extrema. A local max is the highest point in its neighbourhood. The absolute max is the highest point on the entire domain or interval.
Forgetting endpoints in closed-interval problems. The absolute max or min can occur at an endpoint, not just at a critical point.
Thinking that marginal cost equals the cost of one unit. Marginal cost is the cost of one additional unit at a given production level, not the cost of the first unit.
⚠️ Optimisation word problems are high-value exam questions. Practise setting up the function from the word problem, as that is usually the hardest part.
⚠️ Know both the first and second derivative tests. Some exam questions require one; some require the other.
⚠️ Marginal analysis questions often ask you to interpret a derivative value in context. Be ready to write a sentence like: "The cost of producing the 101st unit is approximately $8.50."
⚠️ Inflection points and concavity are tested on the final. Know how to find them and what they mean in context.
True or false: profit is maximised where revenue is maximised. Answer: False. Profit is maximised where marginal revenue equals marginal cost.
Fill in the blank: a function is concave up where f''(x) is ______. Answer: Positive.
True or false: if f'(c) = 0 and f''(c) > 0, then c is a local maximum. Answer: False. f''(c) > 0 means concave up, so c is a local minimum.
On a closed interval, where can an absolute maximum occur? Answer: At a critical point in the interior of the interval, or at one of the endpoints.
Q: The cost function is C(x) = 0.05x² + 10x + 200. Find and interpret C'(100).
A: C'(x) = 0.1x + 10. C'(100) = 0.1(100) + 10 = 20. Producing the 101st unit costs approximately $20.
Q: Find the critical points of f(x) = x³ - 6x² + 9x + 2 and classify them.
A: f'(x) = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3). Critical points at x = 1 and x = 3. f''(x) = 6x - 12. f''(1) = -6 < 0, so x = 1 is a local max. f''(3) = 6 > 0, so x = 3 is a local min.
Q: Find the absolute maximum and minimum of f(x) = x³ - 3x + 1 on [-2, 2].
A: f'(x) = 3x² - 3 = 0 gives x = 1 and x = -1. Evaluate: f(-2) = -8 + 6 + 1 = -1, f(-1) = -1 + 3 + 1 = 3, f(1) = 1 - 3 + 1 = -1, f(2) = 8 - 6 + 1 = 3. Absolute max = 3 (at x = -1 and x = 2). Absolute min = -1 (at x = -2 and x = 1).
Q: A company's revenue is R(x) = 200x - 0.5x² and cost is C(x) = 50x + 1000. Find the production level that maximises profit.
A: P(x) = R(x) - C(x) = -0.5x² + 150x - 1000. P'(x) = -x + 150 = 0, so x = 150. P''(x) = -1 < 0, confirming a maximum.
Graph analysis connects directly to the definite integral: knowing where a function is positive or negative helps you interpret the sign of an integral. The closed interval method for absolute extrema reappears in integral applications. Marginal analysis extends naturally into accumulation problems, where you integrate marginal cost or revenue to find totals.
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