Optimisation problems with differentiation

IB Mathematics: Applications and Interpretation SL· 45 min read

1. Modelling Real-World Problems★★☆☆☆⏱ 15 min

The first and most critical step in any optimisation problem is translating the word problem into a single-variable function you can differentiate. You will almost always get a fixed constraint (e.g. fixed perimeter, fixed volume) that lets you eliminate one variable, leaving an objective function of one independent variable.

📘 Definition

Constraint

A fixed condition given in the problem that relates two variables, allowing you to write your objective function in terms of a single variable.

Example:

A box with a fixed total surface area has surface area as its constraint.

📐 Worked Example

A rectangular garden is to be fenced with 100 m of fencing. One side runs along a house, so no fence is needed there. Model the area of the garden as a function of a single variable.

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1. Define variables: Let $x$ = length of sides perpendicular to the house, $y$ = length of the side parallel to the house.
2
1. Write the fencing constraint:
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$2x + y = 100$
4
1. Rearrange the constraint to isolate $y$ :
5
$y = 100 - 2x$
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1. Write the objective function for area $A$ (the value we want to maximise):
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$A = xy = x(100 - 2x) = 100x - 2x^2$
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1. Define the valid domain: $0 < x < 50$ , since lengths cannot be negative.

Exam tip:

Always confirm the domain of your function after modelling. Invalid (negative) values can lead to nonsensical solutions if you ignore domain bounds.

2. Finding and Classifying Critical Points★★★☆☆⏱ 20 min

Once you have a single-variable objective function, the next step is to find critical points (stationary points where the gradient is zero). These are the only locations where a local maximum or minimum can occur.

For AI SL, the simplest method to classify a critical point is the second derivative test, which tells you if the point is a maximum or minimum.

📘 Definition

Second Derivative Test

For a critical point at $x = a$ : if $f^{''} (a) < 0$ , the point is a local maximum; if $f^{''} (a) > 0$ , it is a local minimum.

📐 Worked Example

For the garden area function $A (x) = 100 x - 2 x^{2}$ , find the critical point and classify it.

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1. Differentiate $A (x)$ to get the first derivative:
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$\frac{dA}{dx} = 100 - 4x$
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1. Set the first derivative equal to zero to find critical points:
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$100 - 4x = 0 \implies x = 25$
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1. Calculate the second derivative:
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$\frac{d^2A}{dx^2} = -4$
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1. Classify the point: since $\frac{d ^{2} A}{d x ^{2}} = - 4 < 0$ , $x = 25$ is a local maximum.

Exam tip:

If the second derivative is zero, use a first derivative sign test to classify the point, but this is very rare for polynomial optimisation problems on AI SL exams.

3. Interpreting Solutions in Context★★★☆☆⏱ 15 min

After classifying the local extremum, you need to confirm it is the global extremum (the absolute maximum or minimum over the entire domain) and report your answer in terms of the original problem.

For the closed domains common in AI SL optimisation problems, the global extremum will always be either at a critical point or at an endpoint of the domain. Always check that your answer makes physical sense.

📐 Worked Example

For the garden problem, find the maximum area and the dimensions that produce it.

1
1. We already confirmed a local maximum at $x = 25$ m.
2
1. Check endpoints of the domain $0 < x < 50$ : at $x = 0$ and $x = 50$ , area $A = 0$ , which is a minimum.
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1. Calculate $y$ from the constraint:
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$y = 100 - 2(25) = 50 \text{ m}$
5
1. Calculate the maximum area:
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$A = 25 \times 50 = 1250 \text{ m}^2$
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1. State the final answer in context: The maximum area of the garden is $1250 m^{2}$ , with dimensions 25 m (perpendicular to the house) and 50 m (parallel to the house).

4. Common Exam Problem Types★★★☆☆⏱ 20 min

Maximising area/volume given a fixed perimeter/surface area
Minimising perimeter/surface area given a fixed area/volume
Minimising construction cost with different costs for different materials
Maximising profit/revenue for price-demand functions

📐 Worked Example

A closed cylindrical can has a fixed volume of $1000 cm^{3}$ . Material for the top and bottom costs 2 cents per $cm^{2}$ , and material for the curved side costs 1 cent per $cm^{2}$ . Find the minimum total cost of the can.

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1. Define variables: $r$ = radius of base, $h$ = height of can
2
1. Volume constraint:
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$V = \pi r^2 h = 1000 \implies h = \frac{1000}{\pi r^2}$
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1. Write total cost function $C$ :
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$C = 2(\pi r^2)(2) + (2\pi r h)(1) = 4\pi r^2 + 2\pi r h$
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1. Substitute $h$ to get $C$ as a function of $r$ :
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$C(r) = 4\pi r^2 + \frac{2000}{r}$
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1. Differentiate and set to zero:
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$C'(r) = 8\pi r - \frac{2000}{r^2} = 0 \implies r^3 = \frac{250}{\pi} \implies r \approx 4.31 \text{ cm}$
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1. Second derivative test: $C^{''} (r) = 8 π + \frac{4000}{r ^{3}} > 0$ for all $r > 0$ , so this is a minimum
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1. Calculate minimum cost: $C \approx 696$ cents = $6.96

5. Common Pitfalls

Wrong move:

Leaving the objective function in two variables before differentiating

Why:

Single-variable differentiation methods taught in AI SL only work for functions of one variable

Correct move:

Always use the given constraint to rearrange and substitute, resulting in a single-variable objective function

Wrong move:

Forgetting to check endpoints of the domain

Why:

The global maximum/minimum can sometimes occur at an endpoint, not a critical point

Correct move:

Always define your domain and check function values at all endpoints to confirm the global extremum

Wrong move:

Missing units in the final answer

Why:

Optimisation problems are real-world, so IB allocates marks for correct units

Correct move:

Always state units for all dimensions, areas, volumes, and costs in your final answer

Wrong move:

Not justifying that the critical point is a maximum/minimum

Why:

IB requires explicit justification to award full marks for optimisation questions

Correct move:

Always use the second derivative test or first derivative sign test to classify your critical point

Wrong move:

Making algebraic errors when rearranging the constraint

Why:

A wrong model will lead to a wrong answer even if differentiation is done correctly

Correct move:

Double-check your substitution step before differentiating, and verify the constraint matches the problem description

6. Quick Reference Cheatsheet

Step	Action	Check
1	Define variables, write constraint	Variables match problem description
2	Eliminate one variable, write objective function	Confirm valid domain of x
3	Differentiate, find critical points	Check differentiation calculation
4	Classify critical point with second derivative test	Confirm it is the required extremum
5	Check endpoints, calculate final value	Add units, answer the question asked

7. Frequently Asked

Do I need to justify my maximum/minimum in IB exams?

Yes, IB examiners require explicit justification that your critical point is the required extremum. You can satisfy this with the second derivative test, a first derivative sign test, or by checking endpoints of the domain.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Box volume optimisation (6 marks)
2023 · 2
Area optimisation (7 marks)
2021 · 1
Cost minimisation (5 marks)

Going deeper

syllabusIB AI SL Calculus Topic GuideOfficial syllabus reference

What's Next

Optimisation with differentiation is a core applied calculus topic that appears frequently on both Paper 1 and Paper 2 of IB AI SL. This sub-topic builds on your understanding of basic differentiation and extrema, and prepares you for more advanced applied calculus topics like related rates, as well as integration applications. The problem-solving skills you learn here (modelling real scenarios with math) are also directly applicable to economics, environmental science, and physics courses you may take at university.