Maxima and Minima

Q: Do I have to justify if a stationary point is a maximum or minimum?

Yes, 4PM1 explicitly requires justification for all stationary point classification, even if the question does not explicitly ask for it. You will lose 1-2 marks per question if you skip this step.

Q: Can I use the gradient sign test instead of the second derivative test?

Both methods are accepted for full marks. The second derivative test is faster for most 4PM1 functions, but use the gradient sign test if the second derivative equals zero at the stationary point.

Q: Do I need to memorise mensuration formulas for practical problems?

Sphere surface area, sphere volume, and curved cone surface area are provided on the 4PM1 formula sheet. You need to recall other formulas such as cylinder volume/surface area, rectangle area, etc.

Edexcel International GCSE Further Pure Mathematics· 9E· 25 min read

1. Justifying Maxima vs Minima★★☆☆☆⏱ 10 min

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📘 Definition

Stationary Point Classification Rules

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 $f (x) = x^{3} - 12 x + 3$

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.

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.

2. Setting Up Practical Optimisation Functions★★★☆☆⏱ 15 min

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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.

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

3. Solving Full Practical Optimisation Problems★★★★☆⏱ 20 min

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

4. Optimisation with Complex Differentiation★★★★☆⏱ 15 min

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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 $200 π$ cm². Find the maximum volume of the can.

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Step 1: Define variables: $r$ = radius, $h$ = height. Surface area constraint: $π r^{2} + 2 π r h = 200 π$
2
Step 2: Rearrange constraint to eliminate $h$ : $r^{2} + 2 r h = 200 ⟹ h = \frac{200 - r ^{2}}{2 r}$
3
Step 3: Write volume objective function
$V(r) = \pi r^2 h = \pi r^2 \times \frac{200 - r^2}{2r} = \frac{\pi}{2}(200r - r^3)$
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Step 4: Differentiate and find stationary point
$\frac{dV}{dr} = \frac{\pi}{2}(200 - 3r^2) = 0 \implies 3r^2 = 200 \implies r = \sqrt{\frac{200}{3}} \approx 8.16 \text{ cm}$
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Step 5: Classify: $\frac{d ^{2} V}{d r ^{2}} = \frac{π}{2} (- 6 r) = - 3 π r < 0$ for all positive $r$ , so this is a maximum.
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Step 6: Compute maximum volume
$V = \frac{\pi}{2}\left(200\times\sqrt{\frac{200}{3}} - \left(\sqrt{\frac{200}{3}}\right)^3\right) \approx 1396 \text{ cm}^3$

5. Common Pitfalls

Wrong move:

Skipping justification of stationary point type

Why:

4PM1 spec explicitly requires justification, you will lose 1-2 marks per question if you omit this step

Correct move:

Always include either the second derivative sign result or gradient sign test result in your solution

Wrong move:

Forgetting to state domain restrictions for practical problems

Why:

Some stationary points may correspond to negative lengths or impossible values, so must be discarded

Correct move:

Write the valid domain for your independent variable immediately after defining your objective function

Wrong move:

Assuming a stationary point with $f^{''} (x) = 0$ is a point of inflection

Why:

A zero second derivative only means the test is inconclusive, the point could still be a maximum or minimum

Correct move:

Use the gradient sign change test to classify points where the second derivative equals zero

Wrong move:

Leaving the final answer as the x-coordinate of the stationary point

Why:

Most exam questions ask for the maximum/minimum value (area, volume etc), not the input value that produces it

Correct move:

Substitute your stationary point x-value back into the original objective function to get the required quantity

Wrong move:

Failing to check domain endpoints for practical problems

Why:

In some cases, the maximum/minimum value may occur at the edge of the domain, not at a stationary point

Correct move:

Compare the function value at the stationary point and at the domain endpoints to confirm your result

6. Quick Reference 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

7. Frequently Asked

Do I have to justify if a stationary point is a maximum or minimum?

Yes, 4PM1 explicitly requires justification for all stationary point classification, even if the question does not explicitly ask for it. You will lose 1-2 marks per question if you skip this step.

Can I use the gradient sign test instead of the second derivative test?

Both methods are accepted for full marks. The second derivative test is faster for most 4PM1 functions, but use the gradient sign test if the second derivative equals zero at the stationary point.

Do I need to memorise mensuration formulas for practical problems?

Sphere surface area, sphere volume, and curved cone surface area are provided on the 4PM1 formula sheet. You need to recall other formulas such as cylinder volume/surface area, rectangle area, etc.

Going deeper

official_documentEdexcel 4PM1 SpecificationOfficial syllabus for 4PM1
study_guideDifferentiation Rules Cheat Sheet

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.