# Maxima and Minima

> Edexcel International GCSE Further Pure Mathematics · 4PM1 2016
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-maxima-and-minima/

This guide covers maxima and minima (optimisation) problems for Edexcel IGCSE Further Pure Math (4PM1), including stationary point justification and practical context problems as required by spec 9E.

**Prerequisites:** [Differentiation rules (power, product, quotient, chain)](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-differentiation-rules/); [Abstract stationary point identification](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-stationary-points/)

## Learning objectives

- Classify stationary points as maxima or minima using valid justifications
- Set up objective functions for practical optimisation problems with domain restrictions
- Apply power, product, quotient and chain differentiation rules to solve maxima/minima problems
- Verify maximum/minimum values by checking domain endpoints where required

## Justifying Maxima vs Minima

**Stationary Point Classification Rules** — 1. Second derivative test: If $f''(x) < 0$ at stationary point, it is a local maximum; if $f''(x) > 0$, it is a local minimum; if $f''(x) = 0$, the test is inconclusive. 2. Gradient sign test: If gradient changes from positive to negative, it is a maximum; negative to positive is a minimum.

**Worked example:** Find and classify the stationary points of $f(x) = x^3 - 12x + 3$

1. Step 1: Compute first derivative and solve for stationary points

   $$f'(x) = 3x^2 - 12 = 0 \implies x^2 =4 \implies x=2, x=-2$$
2. Step 2: Compute second derivative

   $$f''(x) = 6x$$
3. Step 3: Classify each point using second derivative test: For $x=2$, $f''(2)=12>0$, so local minimum. For $x=-2$, $f''(-2)=-12<0$, so local maximum.

> **tip**
>
> Always write your justification clearly, e.g. '$f''(2)=12>0$ so this is a local minimum' to ensure you get full marks for the classification step.

> **Exam tip:** If the second derivative equals zero at the stationary point, always use the gradient sign test instead of assuming it is a point of inflection.

*Calculator:* allowed

## Setting Up Practical Optimisation Functions

For practical problems, you first need to define variables, eliminate extra variables using constraints given in the problem, write the objective function (the quantity you are optimising), and state the valid domain for your independent variable.

**Worked example:** A farmer has 80m of fencing to build a rectangular sheep pen against a wall, so only 3 sides require fencing. Set up the objective function for the area of the pen and state its domain.

1. Step 1: Define variables: Let $x$ = width of pen perpendicular to the wall, $l$ = length parallel to the wall.
2. Step 2: Use fencing constraint: $2x + l = 80 \implies l = 80 - 2x$
3. Step 3: Write objective function (area): $A(x) = x \times l = x(80 - 2x) = 80x - 2x^2$
4. Step 4: State domain: Lengths are positive, so $x>0$ and $80-2x>0 \implies 0 < x < 40$

> **warning**
>
> Never skip the domain step: stationary points outside the valid domain must be discarded, as they are impossible in the problem context.

*Calculator:* allowed

## Solving Full Practical Optimisation Problems

Follow this standard process for all practical optimisation problems: set up the objective function, differentiate, find stationary points, classify them, compute the maximum/minimum value, and check endpoints of the domain to confirm your result.

**Worked example:** Find the maximum area of the sheep pen from the previous example.

1. Step 1: We have $A(x) = 80x - 2x^2$, domain $0 < x < 40$
2. Step 2: Differentiate to find stationary points

   $$\frac{dA}{dx} = 80 - 4x = 0 \implies x = 20$$
3. Step 3: Classify the stationary point: $\frac{d^2A}{dx^2} = -4 < 0$, so this is a local maximum.
4. Step 4: Compute maximum area: $A(20) = 80(20) - 2(20)^2 = 1600 - 800 = 800 \text{ m}^2$
5. Step 5: Check endpoints: $A(0)=0$, $A(40)=0$, so 800 m² is the valid maximum area.

> **Exam tip**
>
> Always give your final answer with the correct units from the problem context, e.g. m² for area, cm³ for volume, to avoid losing easy marks.

*Calculator:* allowed

## Optimisation with Complex Differentiation

For problems involving 3D shapes or non-polynomial functions, you will need to use product, quotient or chain rules to differentiate the objective function. Mensuration formulas given on the formula sheet can be used to set up your function.

**Worked example:** An open-top cylindrical can has a fixed surface area of $200\pi$ cm². Find the maximum volume of the can.

1. Step 1: Define variables: $r$ = radius, $h$ = height. Surface area constraint: $\pi r^2 + 2\pi r h = 200\pi$
2. Step 2: Rearrange constraint to eliminate $h$: $r^2 + 2rh = 200 \implies h = \frac{200 - r^2}{2r}$
3. Step 3: Write volume objective function

   $$V(r) = \pi r^2 h = \pi r^2 \times \frac{200 - r^2}{2r} = \frac{\pi}{2}(200r - r^3)$$
4. Step 4: Differentiate and find stationary point

   $$\frac{dV}{dr} = \frac{\pi}{2}(200 - 3r^2) = 0 \implies 3r^2 = 200 \implies r = \sqrt{\frac{200}{3}} \approx 8.16 \text{ cm}$$
5. Step 5: Classify: $\frac{d^2V}{dr^2} = \frac{\pi}{2}(-6r) = -3\pi r < 0$ for all positive $r$, so this is a maximum.
6. Step 6: Compute maximum volume

   $$V = \frac{\pi}{2}\left(200\times\sqrt{\frac{200}{3}} - \left(\sqrt{\frac{200}{3}}\right)^3\right) \approx 1396 \text{ cm}^3$$

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Skipping justification of stationary point type
  - Why it fails: 4PM1 spec explicitly requires justification, you will lose 1-2 marks per question if you omit this step
  - Correct: Always include either the second derivative sign result or gradient sign test result in your solution
- **Wrong:** Forgetting to state domain restrictions for practical problems
  - Why it fails: Some stationary points may correspond to negative lengths or impossible values, so must be discarded
  - Correct: Write the valid domain for your independent variable immediately after defining your objective function
- **Wrong:** Assuming a stationary point with $f''(x)=0$ is a point of inflection
  - Why it fails: A zero second derivative only means the test is inconclusive, the point could still be a maximum or minimum
  - Correct: Use the gradient sign change test to classify points where the second derivative equals zero
- **Wrong:** Leaving the final answer as the x-coordinate of the stationary point
  - Why it fails: Most exam questions ask for the maximum/minimum value (area, volume etc), not the input value that produces it
  - Correct: Substitute your stationary point x-value back into the original objective function to get the required quantity
- **Wrong:** Failing to check domain endpoints for practical problems
  - Why it fails: In some cases, the maximum/minimum value may occur at the edge of the domain, not at a stationary point
  - Correct: Compare the function value at the stationary point and at the domain endpoints to confirm your result

## Cheatsheet

| Step | Action | Exam Tip |
| --- | --- | --- |
| 1 | Define variables, write objective function | Label variables with units from the problem context |
| 2 | State valid domain for independent variable | Ensure all quantities (length, area etc) are positive |
| 3 | Differentiate objective function to find $f'(x)$ | Use correct differentiation rules, simplify fully |
| 4 | Set $f'(x)=0$, solve for stationary points | Discard any solutions outside your stated domain |
| 5 | Classify stationary points | Use second derivative test first; use gradient test if $f''(x)=0$ |
| 6 | Calculate final value | Substitute x back into objective function, check endpoints, add units |

## What's next

Now that you have mastered maxima and minima for Edexcel IGCSE Further Pure Math (4PM1), you are ready to progress to the next calculus subtopics. Optimisation problems are frequently tested in both Paper 1 and Paper 2 of the 4PM1 exam, so practice past paper questions to build speed and accuracy, especially for setting up objective functions from practical contexts. Remember to always include justification for stationary point classification, as this is one of the most common mark-losing areas for students. If you are still making errors when computing first or second derivatives, revisit the differentiation rules revision guide before moving on to tangents and normals.

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