Teacher page: maximum, minimum, optimisation and inflexion (AA 5.8)

The one thing to do with the figure

Guess the cheapest can before any algebra.

Let them move the slider and call out a radius. Most guess too small, because a tall thin can looks economical and is not.

Then do the calculus and get r = 5, h = 10 exactly, with h = 2r falling out. The exact answer landing on the shape they were circling is what makes the method feel worth having.

f″ = 0 is not sufficient

y = x⁴ at the origin has f″ = 12x², which is zero there and positive on both sides. The concavity never changes, so it is a minimum and not an inflexion.

The guide names this case explicitly. Teach the sign change as part of the definition, not as an afterthought.

The answers

The can	V = 250π gives h = 250/r², S = 2πr² + 500π/r. S′ = 0 at r³ = 125, so r = 5, h = 10, S = 150π ≈ 471.2.
Justification	S″ = 4π + 1000π/r³ > 0 for all positive r, so it is a minimum.
Inflexion types	x³ − 6x² + 9x at x = 2 is non-stationary (gradient −3); x³ at 0 is stationary; x⁴ at 0 is neither.
1. Optimal radius	5.
2. x⁴ at the origin	B, a minimum, because f″ does not change sign.
3. Gradient at the inflexion	−3.

Where the marks go

2 marksForming the function in one variable using the constraint. Most of the question.

1 markDifferentiating and solving.

1 markJustifying maximum or minimum, by either test, on the page.

1 markAnswering what was asked, with units.

What each wrong answer tells you

They give	What it means
10 (Q1)	Gave the height rather than the radius. Read the question back.
"Inflexion" for x⁴ (Q2)	The target error. f″ = 0 without a sign change.
0 (Q3)	Assumed an inflexion must be stationary. The whole point of including this example.
No justification	Common and costly. The test is one line and it is a mark.
r ≈ 5.4	Used a different volume or forgot the 2 in the curved surface area. Check their S.

Other things they will say

"Why is h = 2r?" It falls out of the algebra, and it is a genuinely memorable result: the cheapest closed can is as tall as it is wide. Real cans are not, which is worth thirty seconds about labels and stacking.

"Which test should I use?" Either. The second derivative is quicker; the sign change always works. Use the second derivative unless it comes out zero.

"Do I have to check the endpoints?" If a domain is given, yes. Here r can be any positive number, so there are no endpoints to check.

A possible order

	What is happening
1	The can animation, guessing first.
2	Form S in one variable, slowly, with the constraint written out.
3	Differentiate, solve, justify, answer in a sentence.
4	Inflexions: all three types, with x⁴ as the counterexample.
5	Questions 1 to 3.
6	Set practice with at least one non-stationary inflexion.

Two things not to say

Do not write the function on the board for them. Forming it is the assessed skill and the rest is routine.

Do not let "solve f double dash equals zero" stand as the complete method for an inflexion. Without the sign change it is wrong, and x to the fourth is the proof.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

Stationary pointsFor y = x³ − 6x² + 9x + 1, find them.
y′ = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 at x = 1 and x = 3, where y = 5 and y = 1.
Classify themUse y″ to classify each.
y″ = 6x − 12, so y″(1) = −6, a maximum, and y″(3) = 6, a minimum.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

Find the inflexionFind the point of inflexion of the same curve and the gradient there.
y″ = 0 at x = 2, where y = 3. The gradient is y′(2) = −3, so the curve is falling through it.
A contextA box with a square base of side x and no lid has volume 32 cm³. Show the surface area is x² + 128/x and find the x that minimises it.
Height is 32/x², so area = x² + 4x(32/x²) = x² + 128/x. Differentiating, 2x − 128/x² = 0 gives x³ = 64, so x = 4.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

Justify the minimumFor that box, confirm x = 4 gives a minimum rather than a maximum.
The second derivative is 2 + 256/x³, which is positive for all positive x, so the curve is concave up everywhere and the stationary point must be the minimum. No sign check either side is needed once that is said.
Local is not globalExplain why the maximum at x = 1 on the cubic above is not the largest value the function takes.
A cubic is unbounded, so y grows without limit as x increases. The maximum found is local. Only on a closed interval does a greatest value exist, and then the endpoints are candidates too.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.