A-level · Beta

Applied optimisation

Cut and fold an open box; connect its greatest volume to a zero derivative.

Change the sheet dimensions and corner cut. The net and volume graph update together.

V=x(30−2x)(20−2x)V=x(30-2x)(20-2x)
Change the sheet dimensions and corner cut. The net and volume graph update together.002.5316.895633.787.5950.68101267.6Corner cut x (cm)Volume (cm³)

A slightly larger cut increases volume here.

Box base (cm)
24 × 14
Height (cm)
3
Volume (cm³)
1008
dV/dx (cm²)
108
Best cut (cm)
3.9237
Maximum volume (cm³)
1056.3
Red: remove; dashed: fold.

Use 0 < x < half the shorter sheet side for a physical box. Endpoints show limiting zero-volume cases. Coordinates and volumes are rounded.

Predict, test and explain

Step 1 of 3

Predict

Does the largest possible corner cut give the largest box volume?

InvestigateWhy does the deepest possible box have zero volume?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Try it in the simulation. Change one thing at a time.

Notes and saved values stay in this tab. They disappear when you reload.

Hints, self-check and connections
Try a self-check

Fixed values, separate from the diagram controls.

Does the largest possible corner cut give the largest box volume?

V=x(12−2x)2V=x(12-2x)^2

Hint

Write volume in terms of one variable and check its physical domain.

Reveal answer

Increasing the cut raises the height but shrinks both base dimensions. The boundary has zero volume.

x=2,V=128x=2,\quad V=128

Build an explanation

Hint 1

Write volume in terms of one variable and check its physical domain.

Hint 2

Increasing the cut raises the height but shrinks both base dimensions. The boundary has zero volume.

Hint 3

Use the model to check a prediction, then explain the result.

Worked example

This example uses fixed values, separate from the diagram controls.

Find the best cut for an open box made from a 12 cm by 12 cm sheet.

  1. V = x(12 − 2x)² for 0 < x < 6.
  2. V′ = 144 − 96x + 12x²; solve V′ = 0.
  3. The interior root is x = 2; x = 6 is a degenerate boundary. The derivative changes from positive to negative at 2.

Answer: Cut 2 cm; maximum volume 128 cm³.

Connect this idea

Watch out: Increasing the cut raises the height but shrinks both base dimensions. The boundary has zero volume.

Specification and learning route

AQA G3 · Edexcel Pure 7.3

AS and A-level. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Differentiation; Stationary points; Forming algebraic constraints; Geometry or context formulas.

AQA specification · Edexcel specification