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Find the maximum possible cone volume.

Read the idea, work independently, then explain what changed.

TOPIC 01

2024 JM02

Do not report the squared-volume maximum as a volume.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Find the maximum possible cone volume.

V(x)=π3x21−x2,0≤x≤1V(x)=\frac\pi3x^2\sqrt{1-x^2},\quad0\le x\le1

Official paper · jm02-2024 · 2(a)(v) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 9

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Volume is nonnegative.
Hint 2
Take the positive square root of the maximum of V².
Worked solution
  1. Both endpoint volumes vanish and the interior maximum is larger.

    max⁡V2=4π2/243\max V^2=4\pi^2/243
  2. Recover volume with cubic-metre units.

    Vmax⁡=4π2243=23π27 m3V_{\max}=\sqrt{\frac{4\pi^2}{243}}=\frac{2\sqrt3\pi}{27}\ \mathrm{m}^3

Maximum volume 2√3π/27 m³.

Checks and common pitfalls: Do not report the squared-volume maximum as a volume.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Both endpoint volumes vanish and the interior maximum is larger.
    max⁡V2=4π2/243\max V^2=4\pi^2/243
  • Recover volume with cubic-metre units.
    Vmax⁡=4π2243=23π27 m3V_{\max}=\sqrt{\frac{4\pi^2}{243}}=\frac{2\sqrt3\pi}{27}\ \mathrm{m}^3

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Curriculum and source notes ↗