Area 36: the lowest point
Half the perimeter: x + 36/x
At x = 6: 6 + 6 = 12, so the perimeter is 24
x + 36/x = 12 only at x = 6: the square
AM–GM inequality Why can x + 9/x never be less than 6 when x is positive?
Maths olympiad · 45 minutes
Able students hunt for the rectangle of area 36 with the shortest perimeter, find the pattern on a curve, then name it and prove it in one line. Slides to present, a vote that shows who's stuck, practice that climbs to olympiad proofs, and homework.
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A rectangle has area k and one side x, so half its perimeter is x + k/x. Predict the lowest point for k = 9, then move k. What do the lowest points have in common?
Lowest point: (3, 6)
The idea in one minute
For two positive numbers, the mean is never less than the square root of their product: (a + b) ÷ 2 ≥ √(ab), with equality only when a = b. It is true because a square is never negative: (√a − √b)² ≥ 0 opens out to a + b ≥ 2√(ab). Put a = x and b = k/x, and x + k/x ≥ 2√k, reached at x = √k: of all rectangles with the same area, the square has the shortest perimeter.
The lesson is problem first. Students try rectangles of area 36 before anyone explains, move a curve to see where its lowest point always lies, and only then name the pattern and prove it. Every least value comes with the question 'when is it reached?'. Practice climbs from substituting into the inequality to short written proofs of the kind maths olympiads ask for, which the teacher marks.
By the end, students can
- state AM–GM for two positive numbers, and when the two sides are equal
- prove it from (√a − √b)² ≥ 0
- find least values such as x + k/x and 4x + 9/x, and say where they are reached
- find the greatest product of two positive numbers with a fixed sum
- write a short proof of an inequality, line by line
Key words
- arithmetic mean
- geometric mean
- inequality
- equality case
- least value
- proof
Before this lesson
Students should find a square root, expand (p − q)², and know that a square is never negative. No calculus is needed; it suits able students aged about 13 to 16.
The lesson, step by step
30 minutes in total. Each step says what changes for this length.
3 min 1 · Hook · students think
The shortest fence
Students try rectangles of area 36 before anyone explains.
For this length: Give the five whole-number rectangles at once; vote straight away
Say“A rectangle has area 36 square metres. Which one has the shortest perimeter? Try some, and keep a list. Then: can any rectangle of area 36 beat your best?”
Ask- Do the sides have to be whole numbers?
- Try 4.5 by 8. Is it better or worse than 6 by 6?
Watch forStudents who only try whole-number sides think they have checked every rectangle: ask whether 5 by 7.2 counts.
For the teacher
Two minutes of silent work, listing rectangles and their perimeters. Most start with whole-number sides.
Ask for the best so far, then the real question: is 6 by 6 the best of all, including sides like 5.5 by 6.54…? Trying more cases makes it likely, not certain. The lesson finds the reason.
Collect a few perimeters on the board: 1 by 36 gives 74, 2 by 18 gives 40, 3 by 12 gives 30, 4 by 9 gives 26, 6 by 6 gives 24. Take a vote with Ask the class: can any rectangle of area 36 have a perimeter under 24? Don't settle it: that is the lesson.
Shorter: Give the five whole-number rectangles at once; vote straight away (−2 min).
8 min 2 · Explain · you explain
Discover it, then name it
A curve's lowest point for many areas, then AM–GM named and proved in one line.
For this length: Show the rule from the curve and the proof as one slide
Say“Half the perimeter is x + 36/x. Let's look at every rectangle at once, as a curve, and find its lowest point.”
Ask- Before I move k to 16: where do you think the lowest point will be?
- The lowest points were (2, 4), (3, 6), (4, 8). What is the rule?
- In (√a − √b)² ≥ 0, where did we need a and b positive?
Watch forReading the pattern as 'the least value is 2k': check it with k = 9, where the least value is 6, not 18.
The proof, opened out
(√a − √b)² = a − 2√(ab) + b
A square is never negative, so a + b ≥ 2√(ab)
Equal only when √a = √b, that is a = b
For the teacher
First slide: half the perimeter of a rectangle of area 36 with one side x is x + 36/x. Draw it. The lowest point is at x = 6, height 12: the 6 by 6 square, perimeter 24. This answers the hook, for this area.
Second slide: move k. The lowest points are (1, 2), (2, 4), (3, 6), (4, 8) for k = 1, 4, 9, 16. Students say the rule: the lowest point is at x = √k, at height 2√k. The line y = 2√k touches the curve there and never crosses it.
Third slide: name it. For positive a and b, the arithmetic mean (a + b) ÷ 2 is at least the geometric mean √(ab), equal only when a = b. With a = x and b = k/x, the product is k, so x + k/x ≥ 2√k.
Fourth slide: prove it. (√a − √b)² is a square, so it is 0 or more. Opened out: a − 2√(ab) + b ≥ 0, so a + b ≥ 2√(ab). It is 0 only when √a = √b. A proof is these lines, each following from the one before, with the reason said.
Fifth slide: see it. Four a by b tiles go round the inside of a square of side a + b, leaving a square hole of side a − b. So the big square is at least the four tiles: (a + b)² ≥ 4ab, and there is no hole only when a = b. Square-rooting gives a + b ≥ 2√(ab): the same inequality, as a picture. Ask a student to draw it for 5 by 2 on squared paper.
Vote before each move of k (Ask the class): where is the lowest point for k = 16? Students who find the rule themselves remember it; give it a name only after someone says 'x is √k and the value is 2√k'. The identity check is the proof with b = 4, for x ≥ 0.
Shorter: Show the rule from the curve and the proof as one slide (−4 min).
5 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves.
For this length: Do the first two pairs only
Say“Watch me find a least value and where it is reached. Then you do its twin. Always finish with 'when x = …'.”
Ask- What do the two parts multiply to?
- Which x makes the two parts equal?
Watch forWith a number in front, as in 4x + 9/x, putting equality at x = 3: it is where 4x = 9/x, so x = 1.5.
A least value
x > 0: least value of x + 16/x
x × 16/x = 16, so x + 16/x ≥ 2√16 = 8
Equal when x = 16/x: x = 4
Then theirs
x > 0: least value of x + 25/x, and where
A number in front
x > 0: least value of 4x + 9/x
4x × 9/x = 36, so 4x + 9/x ≥ 2√36 = 12
Equal when 4x = 9/x: x² = 9/4, x = 1.5
Then theirs
x > 0: least value of 9x + 1/x, and where
A fixed sum
a + b = 10, both positive: the greatest ab?
√(ab) ≤ (a + b) ÷ 2 = 5, so ab ≤ 25
Equal when a = b = 5
Then theirs
24 m of fence round a rectangle: the greatest area?
For the teacher
Work the first example aloud, writing each line, and end with where the least value is reached. Students solve its twin while you watch for a missing 'when x = …'.
The second pair catches students who forget the numbers in front: the product 4x × 9/x is 36, and equality is where the two parts are equal. The third pair turns it round: a fixed sum gives a greatest product, which is the hook's question the other way up.
Open the answers slide once most have finished their twins. Check every twin ends with where the value is reached.
Shorter: Do the first two pairs only (−3 min).
9 min 4 · Who's stuck? · students vote
Vote, then climb
One question shows who's stuck; then practice from substituting to short olympiad proofs.
For this length: After the vote, questions 1–5 only
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- What do x and 12/x multiply to?
- Is your answer reached? Which x gives it?
Watch forOne value tried and taken as the least; the 2 in 2√k forgotten; k doubled instead of its square root.
Show your working. For every least or greatest value, say where it is reached. The last questions are proofs: write a line for each step, with its reason.
The question that shows who's stuck
- x is positive. What is the least value of x + 12/x?
In your Classroom, as they answer (an example) - A4
- B10
- C9
- D3
- E2
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- For a = 4 and b = 9: (a + b) ÷ 2 = ___ and √(ab) = ___
- x is positive. The least value of x + 49/x is ___, when x = ___
- x is positive. The least value of 2x + 50/x is ___, when x = ___
- A rectangular pen against a wall has fence on three sides and area 72 m². Its two equal sides are x m. The shortest fence is ___ m, when x = ___
- Ravi says: x + 1/x ≥ 2 for every number x except 0. Is he right?
- a and b are positive and (a + b) ÷ 2 = √(ab). What must be true?
- Prove that a/b + b/a ≥ 2 for all positive a and b, and say when the two sides are equal.
- Stretch: x and y are positive and xy = 12. What is the least value of 3x + 4y?
- Stretch: a, b and c are positive and abc = 1. Prove that (1 + a)(1 + b)(1 + c) ≥ 8. When are the two sides equal?
Support · For the proofs: start from a square that is 0 or more, or from AM–GM for two numbers you choose. Say what you multiply or add, and end with when the two sides are equal.
For the teacher
Everyone chooses before anyone explains. Then ask a student for each answer to say how they got it: what do x and 12/x multiply to, and which x gives their value? Whole numbers tried, a forgotten 2 and a doubled 12 all show up here.
Questions 1–2 are substituting; 3–4 change the look; 5–6 ask why; 7 is a first written proof. 8 and 9 are olympiad level: two letters, and three.
Students write proofs on paper or the board; the quiz keeps their answers for you to mark and comment on.
If the class is split, try a few values of x together on a calculator, some between whole numbers, then ask what the two parts multiply to and which x makes them equal. Long answers (7, 9) are marked by you: give a mark for the right idea, one for each step with its reason, one for the equality case. Model answers are in each question's explanation.
Shorter: After the vote, questions 1–5 only (−6 min).
Longer: Swap and mark: pairs mark each other's proof of question 7, then 9 (+10 min).
5 min 5 · Exit · students show
Least, and where
One least value with its proof and its equality case shows who has the whole idea.
Say“Find the least value, show it can't be less, and say which x gives it.”
Watch forA least value without the x that gives it.
A few lines: the value, why it can't be less, and where it is reached.
- x is positive. Find the least value of 4x + 1/x. Show that it can never be less, and say for which x it is reached.
For the teacher
x is positive. Find the least value of 4x + 1/x.
Show it can never be less, and say for which x it is reached.
Answers with the right value but no 'when' show who has the trick but not the idea: start next time with 'is it reached?'. Set the homework.
Longer: Then: what changes if x may be negative? (+5 min).
One to one: let the student find the best rectangle by trying sides first, then ask them to predict each lowest point before moving k. Ask for every least value 'which x gives it?', and have them say each line of a proof aloud before writing it.
Homework: least values and proofs
For every least or greatest value, say where it is reached. Write proofs a line at a time, each with its reason.
- x is positive. Find the least value of x + 64/x, and the x that gives it.
- A rectangle has area 81 cm². What is the shortest its perimeter can be? Which rectangle gives it?
- Two positive numbers multiply to 50. What is the least their sum can be? Give it exactly and to two decimal places.
- Prove that x² + 1/x² ≥ 2 for every x except 0. When are the two sides equal?
- A rectangular vegetable bed against a wall has an edging on three sides and area 98 m². Its two equal sides are x m. Find the least length of edging, and x.
- a and b are positive and a + b = 1. Prove that 1/a + 1/b ≥ 4. When are the two sides equal?
- Challenge: x is positive. Find the least value of x + 4/(x + 1), and the x that gives it. (Hint: the two parts don't multiply to a number. Can you make two that do?)
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching am–gm inequality: common questions
What is the AM–GM inequality, in one line?
For positive numbers a and b, their mean (a + b) ÷ 2 is never less than √(ab), and the two are equal only when a = b. It follows from (√a − √b)² ≥ 0. The lesson uses the version for two numbers; the same idea works for three or more.
Which competitions is this lesson useful for?
It suits able students aged about 13 to 16 preparing for maths challenges and olympiads, such as the UKMT Intermediate Mathematical Challenge and its follow-on olympiads, or the British Mathematical Olympiad's first round later on. The problems are our own, written in the style of such competitions, not taken from past papers.
My student can find the answer but not write the proof. What helps?
Ask for one step a line, each with its reason ('a square is never negative', 'both are positive'), and always an 'equal when' line at the end. The long questions are marked by you in Quizzes, with a comment, so each student gets feedback on the write-up, not only the number.
Is this too hard for a 13-year-old?
The start is not: it is rectangles and perimeters. The curve and the proof need square roots and expanding (p − q)². Questions 1–5 are within reach of a strong Year 9; questions 8 and 9 are olympiad level, and fine to leave for another day.
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