# Optimisation problems with differentiation

> IB Mathematics: Applications and Interpretation SL · IB AI SL
> Source: https://www.owlsprep.com/study/ib-math-ai-sl-u5-optimisation-problems-with-differentiation/

This module covers solving real-world optimisation problems (maximising area, minimising cost, etc.) using differentiation. You will learn to model problems, find extrema, and justify solutions aligned with IB AI SL exam requirements.

**Prerequisites:** [Differentiation of polynomial functions](https://www.owlsprep.com/study/ib-math-ai-sl-u5-basic-differentiation/); [Finding and classifying critical points](https://www.owlsprep.com/study/ib-math-ai-sl-u5-critical-points-extrema/)

## Learning objectives

- Translate real-world optimisation problems into single-variable mathematical functions
- Identify critical points of objective functions using differentiation
- Classify extrema to confirm maximum/minimum solutions
- Justify and report solutions in the context of the original problem

## Modelling Real-World Problems

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.

**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.

1. 1. Define variables: Let $x$ = length of sides perpendicular to the house, $y$ = length of the side parallel to the house.
2. 2. Write the fencing constraint:
3. $$2x + y = 100$$
4. 3. Rearrange the constraint to isolate $y$:
5. $$y = 100 - 2x$$
6. 4. Write the objective function for area $A$ (the value we want to maximise):
7. $$A = xy = x(100 - 2x) = 100x - 2x^2$$
8. 5. 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.

## Finding and Classifying Critical Points

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.

**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 $A(x) = 100x - 2x^2$, find the critical point and classify it.

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

## Interpreting Solutions in Context

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

> **tip**
>
> Always include correct units in your final answer. IB examiners regularly penalize missing units for optimisation problems.

## Common Exam Problem Types

- 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 $1000 \text{ cm}^3$. Material for the top and bottom costs 2 cents per $\text{cm}^2$, and material for the curved side costs 1 cent per $\text{cm}^2$. Find the minimum total cost of the can.

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

## Common pitfalls

- **Wrong:** Leaving the objective function in two variables before differentiating
  - Why it fails: Single-variable differentiation methods taught in AI SL only work for functions of one variable
  - Correct: Always use the given constraint to rearrange and substitute, resulting in a single-variable objective function
- **Wrong:** Forgetting to check endpoints of the domain
  - Why it fails: The global maximum/minimum can sometimes occur at an endpoint, not a critical point
  - Correct: Always define your domain and check function values at all endpoints to confirm the global extremum
- **Wrong:** Missing units in the final answer
  - Why it fails: Optimisation problems are real-world, so IB allocates marks for correct units
  - Correct: Always state units for all dimensions, areas, volumes, and costs in your final answer
- **Wrong:** Not justifying that the critical point is a maximum/minimum
  - Why it fails: IB requires explicit justification to award full marks for optimisation questions
  - Correct: Always use the second derivative test or first derivative sign test to classify your critical point
- **Wrong:** Making algebraic errors when rearranging the constraint
  - Why it fails: A wrong model will lead to a wrong answer even if differentiation is done correctly
  - Correct: Double-check your substitution step before differentiating, and verify the constraint matches the problem description

## 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 |

## 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.

- [Indefinite integration of polynomials](https://www.owlsprep.com/study/ib-math-ai-sl-u5-indefinite-integration-of-polynomials/)
- [Definite integrals and area under a curve](https://www.owlsprep.com/study/ib-math-ai-sl-u5-definite-integrals-and-area-under/)
- [Kinematic applications of calculus](https://www.owlsprep.com/study/ib-math-ai-sl-u5-kinematic-applications-of-calculus/)

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