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Study GuideYear 12Level 3 · ChallengeHSCVCE

Free Year 12 Applications of Differentiation: Stationary Points and Optimisation

An HSC/VCE-style study guide to using the first and second derivatives to find and classify stationary points and points of inflection, then applying the same method to optimisation problems. Worked examples cover a cubic, a case where the second-derivative test fails, a fencing problem, an open box and a cylinder of fixed volume, with a six-question self-check and answers.

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Year
Year 12
Subject
Mathematics
Topic
Calculus
Difficulty
Level 3 · Challenge
Estimated time
30 minutes
Curriculum
Australian Curriculum
Answers
Not applicable
Format
PDF (A4) + print

Students will practise

  • finding stationary points by solving f′(x) = 0
  • classifying stationary points with the second derivative or a sign table
  • finding points of inflection and intervals where a function is increasing or decreasing
  • setting up and solving optimisation problems, including justifying the maximum or minimum

Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Applications of Differentiation: Stationary Points and Optimisation

  1. 1HSC/VCE-style Calculus Practice with Worked Solutions
  2. 2Year 12 Differentiation (Power Rule) Worksheet — Level 3
  3. 3Year 12 Differentiation (Power Rule) Worksheet — Level 3 (Set 2)
  4. 4Year 12 Integration (Polynomials) Worksheet — Level 2
Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practise with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

Year 12 students (typically ages 17–18) working on calculus. It is pitched at level 3 · challenge.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 30 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try HSC/VCE-style Calculus Practice with Worked Solutions.

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Year 12 · Mathematics · Calculus

Applications of Differentiation: Stationary Points and Optimisation

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What you need to know

The first derivative f′(x) gives the gradient: f′(x) > 0 means the function is increasing, f′(x) < 0 means decreasing, and f′(x) = 0 at a stationary point. The second derivative f″(x) describes concavity: f″(x) > 0 means concave up (shaped like a cup), f″(x) < 0 means concave down (like a cap). A point of inflection is where the concavity changes, so f″(x) = 0 and f″ changes sign there.

At a point where f′(x) = 0Second-derivative testSign of f′ either side
Local minimumf″(x) > 0− then +
Local maximumf″(x) < 0+ then −
Horizontal point of inflectionf″(x) = 0 and f″ changes signsame sign both sides
Test inconclusivef″(x) = 0use the sign table instead

Find and classify the stationary points of y = x3 − 3x2 − 9x + 5, and find the point of inflection

  1. y′ = 3x2 − 6x − 9 = 3(x2 − 2x − 3) = 3(x − 3)(x + 1), so y′ = 0 at x = 3 and x = −1.
  2. y″ = 6x − 6. At x = −1: y″ = −12 < 0, maximum. At x = 3: y″ = 12 > 0, minimum.
  3. y(−1) = −1 − 3 + 9 + 5 = 10 and y(3) = 27 − 27 − 27 + 5 = −22.
  4. Maximum at (−1, 10), minimum at (3, −22).
  5. Inflection: y″ = 0 at x = 1, and y″ changes from negative to positive there. y(1) = 1 − 3 − 9 + 5 = −6, so the point of inflection is (1, −6).

When the second-derivative test fails: y = x4

  1. y′ = 4x3 = 0 at x = 0. y″ = 12x2, and y″(0) = 0, so the test is inconclusive.
  2. Sign table for y′: at x = −1, y′ = −4 (negative); at x = 1, y′ = 4 (positive).
  3. The gradient changes from − to +, so (0, 0) is a minimum, even though y″ = 0 there.

Optimisation method. 1. Draw a diagram and define variables. 2. Write the quantity to be maximised or minimised as a function of one variable, using any constraint to eliminate the others. 3. State the domain. 4. Differentiate, set the derivative to zero and solve. 5. Justify that it is a maximum or minimum (second derivative or sign table). 6. Answer the actual question, with units, and check the endpoints of the domain if there are any.

A farmer has 100 m of fencing for a rectangular paddock against a straight river. No fence is needed along the river. Find the maximum area.

  1. Let the two sides perpendicular to the river be x m and the side parallel be y m. Fence used: 2x + y = 100, so y = 100 − 2x (with 0 < x < 50).
  2. Area: A = xy = x(100 − 2x) = 100x − 2x2
  3. A′ = 100 − 4x = 0 gives x = 25. A″ = −4 < 0, so this is a maximum.
  4. y = 100 − 50 = 50. Maximum area = 25 × 50 = 1250 m[sup 2].

Squares of side x cm are cut from the corners of a 24 cm × 24 cm sheet, which is folded into an open box. Find the x that gives the largest volume.

  1. V = x(24 − 2x)2, for 0 < x < 12.
  2. Product rule: V′ = (24 − 2x)2 + x × 2(24 − 2x)(−2) = (24 − 2x)[(24 − 2x) − 4x] = (24 − 2x)(24 − 6x)
  3. V′ = 0 gives x = 12 (volume zero, reject) or x = 4.
  4. Sign check: V′(3) = 18 × 6 > 0 and V′(5) = 14 × (−6) < 0, so x = 4 is a maximum.
  5. Maximum volume: V(4) = 4 × 162 = 1024 cm[sup 3] when x = 4 cm.

A closed cylindrical can must hold 500 cm3. Find the radius that minimises the surface area.

  1. Constraint: πr2h = 500, so h = .
  2. Surface area: S = 2πr2 + 2πrh = 2πr2 + 2πr × = 2πr2 +
  3. S′ = 4πr − = 0, so 4πr3 = 1000 and r3 = = 79.58…, giving r = 4.30 cm (to 2 decimal places).
  4. S″ = 4π + > 0 for all r > 0, so this is a minimum.
  5. r ≈ 4.30 cm (and then h = ≈ 8.60 cm, which is 2r — the most efficient can is as tall as it is wide).
Common mistake: stopping at the x-value. If the question asks for the maximum area, volume or profit, substitute back to find that value, include units, and make sure your answer lies inside the domain. If the domain is a closed interval, also check the endpoint values — the overall maximum might be there.
Common mistake: claiming a maximum or minimum without justification. Examiners expect a reason: either the sign of the second derivative at that point or a sign table of the first derivative on either side.

Check yourself

  1. Find the stationary point of y = x2 − 6x + 1.
  2. Use the second derivative to classify the stationary point in question 1.
  3. Find the x-values of the stationary points of y = 2x3 − 3x2 − 12x.
  4. Classify each stationary point in question 3.
  5. Find the x-value of the point of inflection of y = 2x3 − 3x2 − 12x.
  6. Two positive numbers add to 20. Find the numbers that give the largest product, and that product.

Answers: 1. y′ = 2x − 6 = 0 at x = 3; y = 9 − 18 + 1 = −8, so (3, −8) 2. y″ = 2 > 0, minimum 3. y′ = 6x2 − 6x − 12 = 6(x − 2)(x + 1), so x = 2 and x = −1 4. y″ = 12x − 6: at x = 2, y″ = 18 > 0 (minimum); at x = −1, y″ = −18 < 0 (maximum) 5. 12x − 6 = 0, so x = 6. P = x(20 − x), P′ = 20 − 2x = 0 at x = 10, P″ = −2 < 0; the numbers are 10 and 10, product 100

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Last reviewed
1 October 2026
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