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Find the minimum of ab under the positive-variable condition.

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

TOPIC 01

2021 JM01

Equality requires 5/a=4/b, not a=b.

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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 minimum of ab under the positive-variable condition.

5/a+4/b=3,a,b>05/a+4/b=3,\quad a,b>0

Official paper · jm01-2021 · I.15 · PDF 3

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A20/920/9
  2. Option B20/320/3
  3. Option C80/980/9
  4. Option D80/380/3
  5. Option E20/3\sqrt{20}/3

Working and explanation

BUILD THE REASONING

Hint 1
Let u=5/a and v=4/b; their sum is fixed.
Hint 2
Maximize uv to minimize ab=20/(uv).
Worked solution
  1. Apply AM–GM to the two positive new variables.

    u+v=3⇒uv≤(3/2)2=9/4u+v=3\Rightarrow uv\le(3/2)^2=9/4
  2. Invert the positive bound and verify equality.

    ab=20/(uv)≥80/9;a=10/3, b=8/3ab=20/(uv)\ge80/9;\quad a=10/3,\ b=8/3

C: 80/9.

Checks and common pitfalls: Equality requires 5/a=4/b, not a=b.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Apply AM–GM to the two positive new variables.
    u+v=3⇒uv≤(3/2)2=9/4u+v=3\Rightarrow uv\le(3/2)^2=9/4
  • Invert the positive bound and verify equality.
    ab=20/(uv)≥80/9;a=10/3, b=8/3ab=20/(uv)\ge80/9;\quad a=10/3,\ b=8/3

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Curriculum and source notes ↗