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Maximise the linear objective over the feasible region.

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

TOPIC 01

2025 JM01

An upper bound is a maximum only after a feasible equality case is found.

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

01 / Standard#Your turn

Maximise the linear objective over the feasible region.

3x+4y≤7,x−2y≥−1,y≥−1,z=3x+y3x+4y\le7,\quad x-2y\ge-1,\quad y\ge-1,\quad z=3x+y

Official paper · jm01-2025 · I.14 · PDF 3

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

Skills and prerequisite lessons
  1. Option A44
  2. Option B66
  3. Option C77
  4. Option D1010
  5. Option E1111

Working and explanation

BUILD THE REASONING

Hint 1
Use the first constraint to bound 3x+y.
Hint 2
Then use y≥−1 and test attainability.
Worked solution
  1. Eliminate x from an upper bound for the objective.

    3x+y≤7−3y≤103x+y\le7-3y\le10
  2. Both bounds attain equality at a feasible point.

    (x,y)=(113,−1),x−2y=173≥−1(x,y)=\left(\frac{11}3,-1\right),\quad x-2y=\frac{17}3\ge-1

D: maximum 10.

Checks and common pitfalls: An upper bound is a maximum only after a feasible equality case is found.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Eliminate x from an upper bound for the objective.
    3x+y≤7−3y≤103x+y\le7-3y\le10
  • Both bounds attain equality at a feasible point.
    (x,y)=(113,−1),x−2y=173≥−1(x,y)=\left(\frac{11}3,-1\right),\quad x-2y=\frac{17}3\ge-1

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Curriculum and source notes ↗