Teacher page: optimisation in context

Kinematics is not set at Standard Level on this course. Displacement, velocity and acceleration belong to the Higher Level list. Do not use a moving object as an optimisation context for an SL class, and tell them it is not revision they need.

The one thing to do with the animation

Run it before any algebra. Ask what happens to the box as the corner squares get bigger. Most will say it gets taller. Fewer will say the base gets smaller at the same time, and that competition is the entire problem.

Then ask where the best box is, by eye, from the volume graph. They will get close to 2. Calculus then earns its place by giving the exact answer rather than a good guess, which is a better motivation than being told to differentiate.

The answers

The box	V = x(12 − 2x)², 0 < x < 6. V′ = (12 − 2x)(12 − 6x), zero at x = 2 and x = 6. Maximum 128 cm³ at x = 2, an 8 by 8 base 2 cm deep. For comparison V(1) = 100 and V(3) = 108.
The fence	A = x(40 − 2x), A′ = 40 − 4x, x = 10. 200 m², a 10 by 20 pen.
1. 20 cm sheet, x = 3	588 cm³. Base 14 by 14, height 3.
2. Why reject x = 6	B. The base would be 12 − 12 = 0.
3. 60 m of fencing	450 m², from x = 15 and a 15 by 30 pen.

Where the marks go

2 marksForming the function in one variable, usually with a substitution from a constraint. This is most of the question and most of the difficulty.

1 markDifferentiating and solving.

1 markRejecting the impossible root with a reason. The reason is the mark, not the rejection.

1 markAnswering the question asked, in context, with units. If it asks for the volume, x = 2 is not an answer.

What each wrong answer tells you

They write	What it means
1200 (Q1)	Used the full 20 by 20 as the base. They have not seen that cutting x from BOTH ends takes 2x off the side.
196 (Q1)	Base area only. Forgot to multiply by the height. Common when the height is the same letter as the cut.
"6 is bigger than 2" (Q2)	No contextual reasoning at all. Push for what physically happens to the card.
15 or 30 (Q3)	Answered with a length when an area was asked for. The commonest end-of-question error across the whole sub-topic.
900 (Q3)	Treated all four sides as fenced, ignoring the wall. Draw the pen before writing anything.

Other things they will say

"Do I have to prove it is a maximum?" If the question says show or justify, yes, and a sign check either side of x = 2 is enough. Often the context makes it obvious and the marks are elsewhere; read the command term.

"Can I just use the calculator's maximum?" For a check, yes, and encourage it. As the method it throws away the setup and derivative marks, which are most of the question.

"The wall side is always double." True for this family of fencing problems and worth noticing, but a pattern is not a method. Change the numbers and make them redo it.

A possible order

	What is happening
1	The animation with no algebra. What competes with what? Guess the best x from the graph.
2	Build V = x(12 − 2x)² together, slowly, including the range. Do not rush to differentiate.
3	Differentiate, solve, reject x = 6 out loud with a reason, answer in a sentence.
4	The fence problem, with them forming the function first and you checking before anyone differentiates.
5	Questions 1 to 3. Question 3 catches anyone answering with a length.
6	The habit: underline what the question actually asks for before writing the final line.

Two things not to say

Do not hand out the function. If you write V = x(12 − 2x)² on the board, you have done the part worth two marks and the lesson becomes an exercise in differentiating a bracket.

Do not say "ignore the other root". Name it, reject it, and say why in a full sentence. Students copy the phrasing you model, and a mark depends on it.

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.

Evaluate the modelA 20 cm by 20 cm sheet has squares of side x cut from the corners. Write the volume and find it at x = 3.
V = x(20 − 2x)², so V(3) = 3(14)² = 588 cm³.
Differentiate and solveFind the stationary values of that volume.
V′ = (20 − 2x)(20 − 6x) = 0 at x = 10 and x = 10/3.

2In context

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

Reject and concludeWhich root is valid, and what is the maximum volume?
x = 10 makes the base side 20 − 20 = 0, so there is no box. x = 10/3 = 3.33 cm gives V = 592.59 cm³.
The 12 cm versionFor a 12 cm sheet, V′ = 0 at x = 2 and x = 6. State the maximum volume and why x = 6 is rejected.
V(2) = 2(8)² = 128 cm³. At x = 6 the base side is 12 − 12 = 0, so the box has no volume.

3Challenge

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

State the domain firstExplain why writing down 0 < x < 10 before differentiating prevents the usual error.
It makes the invalid root visible before it is found, rather than being spotted afterwards or not at all. The physical limits of x are part of the model, not an afterthought.
Justify the maximumYou found x = 10/3 gives a stationary point. Show it is a maximum without the second derivative.
V = x(20 − 2x)² is zero at x = 0 and x = 10 and positive between, so the single interior stationary point must be the maximum. The endpoints bounding a positive continuous function do the work.

Practicalities

The sheet and the volume graph are one figure, so they stay in step on a phone. No external library, nothing stored, nothing sent.