Source: Calc AB Lecture Notes, University of Central Florida
Tags: optimisation, optimization, applied max min problems, primary equation, secondary equation, constraint equation, Mean Value Theorem, MVT, average rate of change, instantaneous rate of change, AP Calculus AB Unit 3
Difficulty: Intermediate | Prerequisites: Differentiation rules, critical numbers, first and second derivative tests (see Part 1 of these notes).
Optimisation is the payoff for everything you have learned about derivatives so far. Instead of abstract functions, you are working with real quantities: area, volume, cost, distance. The setup is always the same: translate a word problem into equations, reduce to one variable, then use a derivative test to find the max or min. The Mean Value Theorem, covered at the end of the unit, is more theoretical but appears regularly on the AP exam as a justification tool. Both topics assume you are comfortable finding and classifying critical numbers.
Optimisation problems ask you to maximise or minimise a real-world quantity by setting up two equations (a primary equation for the quantity you care about, and a secondary/constraint equation to eliminate a variable), then using calculus to find the critical number that gives the best answer. The Mean Value Theorem guarantees that a smooth function achieves its average rate of change as an actual instantaneous rate of change at some interior point.
Optimisation (optimization)
The process of finding the maximum or minimum value of a quantity in an applied context. In simple terms, you are answering "what is the biggest/smallest this thing can be?"
Primary equation
The equation representing the quantity you want to maximise or minimise (area, volume, profit, cost, distance, etc.). This is what you will eventually differentiate.
Secondary equation (constraint equation)
An equation that relates the variables in the problem and reflects a given restriction (a fixed perimeter, a budget limit, a sum that equals a constant). You use this to eliminate one variable from the primary equation so it depends on only one variable.
Mean Value Theorem (MVT)
If f is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in (a, b) where f'(c) = (f(b) – f(a)) / (b – a). Think of it as: at some point, the instantaneous speed exactly equals the average speed over the trip.
Average rate of change
The slope of the secant line between two points on a graph: (f(b) – f(a)) / (b – a). This is the "overall" rate across an interval.
Instantaneous rate of change
The derivative f'(c) at a single point. This is the slope of the tangent line at that point.
Read the problem and identify what to maximise or minimise. This gives you your primary equation. Label it clearly (e.g., A = area, V = volume, P = product).
Identify the constraint. There will be a fixed total, a boundary, or some relationship between your variables. Write it as an equation.
Use the constraint to eliminate one variable. Solve the secondary equation for one variable, then substitute into the primary equation so it is in terms of a single variable.
Differentiate the primary equation with respect to the remaining variable.
Set the derivative equal to zero and solve for the critical number(s).
Verify that the critical number gives a max or min (using the first or second derivative test, or by reasoning about the domain).
Answer the question asked. The problem may want both variables, not just the one you solved for.
Problem: Find two numbers that add up to 20 whose product is as large as possible.
Primary equation: P = xy (the product you want to maximise).
Secondary equation: x + y = 20.
Eliminate x: x = 20 – y. Substitute: P = (20 – y)(y) = 20y – y².
Differentiate: dP/dy = 20 – 2y.
Set equal to zero: 20 – 2y = 0, so y = 10.
Then x = 20 – 10 = 10.
The maximum product is 10 × 10 = 100. Both numbers are 10.
Problem: A rectangle is bounded by the x-axis, the y-axis, and the graph of y = (6 – x)/2. Find the dimensions that maximise the area.
The rectangle has one corner at the origin, width x, and height y = (6 – x)/2 = –(1/2)x + 3.
Primary equation: A = x · y = x(–(1/2)x + 3) = –(1/2)x² + 3x.
Differentiate: dA/dx = –x + 3.
Set equal to zero: –x + 3 = 0, so x = 3.
Then y = –(1/2)(3) + 3 = 3/2.
Maximum area dimensions: x = 3 and y = 3/2. The maximum area is 3 × 3/2 = 9/2.
Maximising the area of a rectangle inscribed in a curve or under a line.
Minimising the surface area of a box or can for a given volume.
Minimising the distance from a point to a curve.
Maximising revenue or profit given a demand function.
In every case, the method is the same: primary equation, constraint, substitute, differentiate, solve, verify, answer.
Statement: If f is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in (a, b) such that:
f'(c) = (f(b) – f(a)) / (b – a)
What it means in plain language: Over any smooth journey, there is at least one moment where your instantaneous speed equals your average speed for the whole trip.
How to apply it:
Verify the hypotheses: f must be continuous on the closed interval and differentiable on the open interval.
Compute the average rate of change: (f(b) – f(a)) / (b – a).
Set f'(c) equal to that value and solve for c.
Confirm that c lies in (a, b).
Relationship to Rolle's Theorem: Rolle's Theorem is the special case of MVT where f(a) = f(b), which forces the average rate of change to be zero, guaranteeing a point where f'(c) = 0.
Optimisation template:
Primary equation: the quantity to optimise (express in terms of two variables).
Constraint: the fixed relationship between the variables.
After substitution, differentiate the single-variable expression, set the derivative to zero, and solve.
Mean Value Theorem formula:
f'(c) = (f(b) – f(a)) / (b – a)
where c is in (a, b), f is continuous on [a, b], and f is differentiable on (a, b).
Optimisation is everywhere in engineering and business. Packaging companies minimise cardboard for a given box volume. Farmers maximise enclosed area with a fixed length of fencing. Economists find the production quantity that maximises profit. The Mean Value Theorem underpins speed-camera enforcement logic: if you cover a stretch of motorway faster than the speed limit allows, MVT guarantees you exceeded the limit at some point, even if no single camera caught you speeding.
Students jump straight to differentiating without properly setting up the primary and secondary equations. The setup is usually where marks are lost, not the calculus.
Students forget to substitute the constraint before differentiating. If your expression still has two variables when you take the derivative, something has gone wrong.
Students find the critical number and stop, forgetting to answer the original question. If the problem asks for dimensions, you need both x and y, not just one.
For MVT, students often forget to verify the hypotheses. The AP exam expects you to state that f is continuous on [a, b] and differentiable on (a, b) before applying the theorem.
⚠️ Optimisation problems appear on the AP free-response section regularly. Full marks require a clear setup (equations labelled), showing the derivative work, and verifying the result is a max or min.
⚠️ The Mean Value Theorem is a favourite justification question. Typical prompt: "Explain why there must exist a value c where f'(c) = [some value]." Your answer must cite MVT by name and verify both hypotheses.
⚠️ Do not confuse the Mean Value Theorem with the Intermediate Value Theorem (IVT). MVT is about derivatives and rates of change. IVT is about function values and continuity.
⚠️ On multiple-choice questions, wrong answers often come from skipping the substitution step or solving for the wrong variable.
True or false: In an optimisation problem, the primary equation is the constraint.
Fill in the blank: Before differentiating the primary equation, you must use the __________ equation to reduce to one variable.
True or false: The Mean Value Theorem requires f to be differentiable on the closed interval [a, b].
Fill in the blank: Rolle's Theorem is a special case of MVT where f(a) = __________.
True or false: If two numbers add to 20 and you want to maximise their product, the answer is 10 and 10.
Answers: 1. False (the primary equation is what you optimise; the constraint is separate). 2. Secondary / constraint. 3. False (differentiable on the open interval (a, b), continuous on the closed interval [a, b]). 4. f(b). 5. True.
Q: Outline the general steps for solving an optimisation problem.
A: Identify the quantity to maximise or minimise and write it as the primary equation. Write the constraint equation. Use the constraint to eliminate one variable from the primary equation. Differentiate, set equal to zero, solve for the critical number. Verify it is a max or min. Solve for any remaining variables and answer the question asked.
Q: Two positive numbers sum to 20. What is their maximum product, and what are the numbers?
A: Let the numbers be x and y with x + y = 20. The product P = x(20 – x) = 20x – x². dP/dx = 20 – 2x = 0 gives x = 10, so y = 10. The maximum product is 100.
Q: A rectangle is bounded by the axes and the line y = (6 – x)/2. Find the dimensions that maximise area.
A: Area A = x · (–(1/2)x + 3) = –(1/2)x² + 3x. dA/dx = –x + 3 = 0 gives x = 3. Then y = 3/2. The dimensions are 3 by 3/2, giving maximum area 9/2.
Q: State the Mean Value Theorem and explain its geometric meaning.
A: If f is continuous on [a, b] and differentiable on (a, b), there exists c in (a, b) with f'(c) = (f(b) – f(a))/(b – a). Geometrically, this means there is a point on the curve where the tangent line is parallel to the secant line connecting (a, f(a)) and (b, f(b)).
Q: How does Rolle's Theorem follow from the Mean Value Theorem?
A: Rolle's Theorem is MVT with the additional condition f(a) = f(b). This makes (f(b) – f(a))/(b – a) = 0, so MVT guarantees a c in (a, b) where f'(c) = 0.
Optimisation connects to related rates (both use derivatives to solve applied problems, but related rates involve time-dependent quantities). The Mean Value Theorem connects to the Fundamental Theorem of Calculus: the FTC can be viewed as an integral version of MVT. In multivariable calculus, optimisation extends to Lagrange multipliers, which handle constraints in higher dimensions.
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