# Optimization applications

> IB Mathematics: Analysis and Approaches SL · IB AA SL Calculus
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u5-optimization-applications/

This sub-topic teaches you to apply differentiation to solve real-world optimization problems, where you find the maximum or minimum value of a quantity subject to given constraints, a frequently tested exam topic for IB AA SL.

**Prerequisites:** [Differentiation of polynomial and rational functions](https://www.owlsprep.com/study/ib-math-aa-sl-u4-differentiation-rules/); [Finding and classifying stationary points](https://www.owlsprep.com/study/ib-math-aa-sl-u5-stationary-points/)

## Learning objectives

- Set up optimization problems from real-world contexts
- Find critical points of an objective function
- Identify and justify global maximum/minimum values
- Solve common applied optimization problems in geometry and business

## The General Optimization Method

All optimization problems follow a consistent framework that helps you avoid errors and earn full method marks in IB exams.

**Optimization Problem** — A problem that requires finding the maximum or minimum value of an objective function, subject to one or more constraints that limit the possible values of the problem variables.

1. Identify the quantity to optimize (maximize/minimize) as your objective function
2. List all given constraints on the variables in the problem
3. Rewrite the objective function as a function of a single variable using the constraints
4. Differentiate the single-variable function to find the first derivative
5. Set the first derivative equal to zero and solve for critical points
6. Check that the critical point is in the domain of the function, then classify it
7. Justify that the critical point is a global extremum (required for full marks)

**Worked example:** A rectangular garden is built alongside a house, so no fencing is needed along the house. You have 100 m of fencing total. Find the dimensions that give the maximum area of the garden.

1. Let $x$ = length of sides perpendicular to the house, $y$ = length parallel to the house. We want to maximize area $A = xy$.
2. The total fencing constraint is: $2x + y = 100$, so rearrange to get $y = 100 - 2x$. Domain is $0 < x < 50$.
3. Substitute into the area formula to get a single-variable objective function:
4. $$A(x) = x(100 - 2x) = 100x - 2x^2$$
5. Differentiate and set equal to zero:
6. $$\frac{dA}{dx} = 100 - 4x = 0 \implies x = 25$$
7. Classify and justify: The second derivative is $A''(x) = -4 < 0$, so this is a local maximum. As it is the only critical point in the domain, it is the global maximum.
8. Find $y = 100 - 2(25) = 50$. The maximum area dimensions are 25 m (perpendicular) × 50 m (parallel).

> **Exam tip:** Always explicitly state the domain of your objective function and justify your global extremum to earn all method marks.

*Calculator:* allowed

IB exams most frequently test optimization with geometric constraints, including fixed perimeter, area, or volume. These follow predictable patterns and can be solved with the general framework.

**Worked example:** An open-top box is made by cutting equal squares of side length $x$ from each corner of a 20 cm × 30 cm cardboard sheet, then folding up the sides. Find $x$ that maximizes the box volume.

1. After folding, the box has height $x$, length $30 - 2x$, width $20 - 2x$. We want to maximize volume $V = x(30-2x)(20-2x)$.
2. Domain: All side lengths are positive, so $0 < x < 10$.
3. $$V(x) = 4x^3 - 100x^2 + 600x$$
4. Differentiate, set to zero:
5. $$V'(x) = 12x^2 - 200x + 600 = 0$$
6. Solve the quadratic: Solutions are $x \approx 12.75$ (outside domain, reject) and $x \approx 3.92$ cm.
7. Second derivative $V''(x) = 24x - 200$, so $V''(3.92) \approx -106 < 0$, so this is a global maximum.
8. Final answer: Maximum volume when $x \approx 3.9$ cm (3 significant figures).

*Calculator:* allowed

IB exams require explicit justification that your critical point gives a global (not just local) maximum or minimum. There are three accepted methods for justification.

- For continuous functions on closed intervals: Evaluate at critical points and endpoints, the largest/smallest value is global
- For open domains: If there is only one critical point that is a local maximum/minimum, it is global
- First derivative test: Show the function changes from increasing to decreasing (max) or vice versa (min) across the critical point

**Worked example:** A company's total cost for $x$ units is $C(x) = 0.1x^2 + 10x + 900$, and revenue is $R(x) = 50x$. Find the number of units that maximizes profit, and justify your answer.

1. Profit is $P(x) = R(x) - C(x)$, so:
2. $$P(x) = -0.1x^2 + 40x - 900$$
3. Domain: $x > 0$, and profit is positive for $0 < x < 373$.
4. Differentiate and find critical point:
5. $$P'(x) = -0.2x + 40 = 0 \implies x = 200$$
6. Justification: This is the only critical point in the domain. The second derivative is $P''(x) = -0.2 < 0$, so it is a local maximum. Therefore it is the global maximum.
7. Final answer: Profit is maximized at 200 units.

> **Exam tip:** You can earn justification marks even if you use a calculator to find the critical point, as long as you state your reasoning clearly.

*Calculator:* allowed

Some optimization problems produce non-polynomial objective functions, but the same general method still applies. You only need to use the correct differentiation rule for the function type.

**Worked example:** A closed cylindrical can must have a volume of 1000 cm³. Find the radius and height that minimize the total surface area of the can.

1. Let $r$ = radius, $h$ = height. We want to minimize surface area $S = 2\pi r^2 + 2\pi r h$. Constraint: Volume $\pi r^2 h = 1000$, so $h = \frac{1000}{\pi r^2}$.
2. Substitute to get objective function, domain $r > 0$:
3. $$S(r) = 2\pi r^2 + \frac{2000}{r}$$
4. Differentiate and set equal to zero:
5. $$S'(r) = 4\pi r - \frac{2000}{r^2} = 0 \implies 4\pi r^3 = 2000 \implies r^3 = \frac{500}{\pi}$$
6. Solve for $r \approx 5.42$ cm. Second derivative:
7. $$S''(r) = 4\pi + \frac{4000}{r^3} > 0 \text{ for all } r > 0$$
8. Justification: Only one critical point, and it is a local minimum, so it is global. Find $h = 2r \approx 10.84$ cm.

> **tip**
>
> For a fixed volume closed cylinder, the minimum surface area always occurs when $h = 2r$, you can use this to check your answers quickly.

*Calculator:* allowed

### Common pitfalls

- **Wrong:** Accepting critical points outside the problem's valid domain
  - Why it fails: Many quadratic equations produce two roots, but one is often outside the domain, leading to an incorrect answer
  - Correct: Always define the domain immediately after writing your single-variable objective function, and reject any out-of-domain critical points
- **Wrong:** Mixing up the objective function and the constraint
  - Why it fails: You end up optimizing the wrong quantity, leading to zero marks for the question even if algebra is correct
  - Correct: Explicitly state what you are optimizing and what your constraint is before starting any calculations
- **Wrong:** Forgetting to justify the extremum is global
  - Why it fails: IB markers require explicit justification to award full marks, even if your final value is correct
  - Correct: Add one clear sentence justifying why the critical point is the global maximum or minimum, using one of the accepted methods
- **Wrong:** Ignoring endpoints when the domain is closed
  - Why it fails: The global extremum can occur at an endpoint of a closed interval, even if there are no critical points inside
  - Correct: For closed intervals, always evaluate the objective function at all critical points and endpoints to find the global extremum
- **Wrong:** Making algebraic errors during substitution
  - Why it fails: A wrong objective function leads to a wrong final answer, even if all subsequent steps are correct
  - Correct: Double-check your substitution and simplification of the objective function before differentiating

### Cheatsheet

| Step | Action |
| --- | --- |
| 1 | Define objective function and list constraints |
| 2 | Rewrite objective as a single-variable function |
| 3 | State the valid domain of the function |
| 4 | Differentiate, solve $f'(x) = 0$ for critical points |
| 5 | Reject critical points outside the domain |
| 6 | Classify the remaining critical point |
| 7 | Justify it is the global extremum |
| 8 | Calculate the requested final value |

### What's next

Optimization is one of the most practical applications of differential calculus, used widely in engineering, business, economics, and data science to make the best possible decisions under resource constraints. Mastering this sub-topic also reinforces your core understanding of differentiation and stationary points, which are foundational to all calculus topics in IB AA SL. After mastering basic optimization applications covered here, you can move on to more complex optimization problems that involve trigonometric functions or appear alongside related rates, which sometimes feature in longer Paper 2 questions. Optimization also builds the foundation for understanding integral calculus applications, where you solve problems involving accumulation of quantities. Regular practice of different problem types will ensure you can earn full marks on this frequently tested topic.

- [Finding and classifying stationary points](https://www.owlsprep.com/study/ib-math-aa-sl-u5-stationary-points/)
- [Differentiation rules for common functions](https://www.owlsprep.com/study/ib-math-aa-sl-u4-differentiation-rules/)
- [Related rates applications](https://www.owlsprep.com/study/ib-math-aa-sl-u6-related-rates/)

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