LEARN · EXPLAIN · REVISE

For integer x,y≥0 and 2x+2y≤2t+1, find max(x+y).

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

T04高二理組數學思維本(2026).pdf · III. Simple linear programming · PDF 11 / printed page 10

TOPIC 01

Linear programming and discrete feasibility

Draw a feasible region and optimise a linear objective, then check integer restrictions when required.

Draw a feasible region and optimise a linear objective, then check integer restrictions when required.
Justify the method and check the conditions in a new situation.

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

01 / Standard#Your turn

t=11

A vertex search needs a nonempty bounded polygon or a justified finite optimum; integer solutions may differ from continuous optima.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Write the defining equation, retain exclusions, and then solve or compare.

Hint 2

Use this intermediate relation.

x+y\le t+1/2;\quad x+y\in\mathbb Z

Worked solution

Write the defining equation, retain exclusions, and then solve or compare.
Apply the stated relation and retain its conditions.
$x+y\le11\text{ and }(11,0)\text{ is feasible}$
The integer optimum differs from the continuous boundary value t+1/2.

The requested value is 11.

Checks and common pitfalls: The integer optimum differs from the continuous boundary value t+1/2.

Think first. Reveal a hint when the class is ready.

Teacher preparation and assessment

Draw a feasible region and optimise a linear objective, then check integer restrictions when required.
Which condition is essential in linear programming and discrete feasibility?
Must rounding a continuous optimum preserve feasibility?

Model or definition: Draw a feasible region and optimise a linear objective, then check integer restrictions when required.
$z=ax+by$
Conditions: A vertex search needs a nonempty bounded polygon or a justified finite optimum; integer solutions may differ from continuous optima.

An objective has no optimum if it is unbounded in an improving feasible direction.