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Profit is P(x)=−x²+2tx−3, for 0≤x≤t+2. Find maximum profit.

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

高二選擇性必修第二册(A版).pdf · 5.3 · PDF 89 / printed page 84

TOPIC 01

Applications of derivatives

Use derivative signs, boundary values and models to justify extrema and optimization.

Use derivative signs, boundary values and models to justify extrema and optimization.
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 / Transfer#Your turn

t=14

A critical point needs a sign or second-derivative check; closed-interval global extrema also require endpoints.

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Differentiate the specified function and substitute only after differentiating.

Hint 2

Use this intermediate relation.

P\prime(x)=2t-2x

Worked solution

Differentiate the specified function and substitute only after differentiating.
Apply the stated relation and retain its conditions.
$x=14\text{ lies in the domain}$
Apply the stated relation and retain its conditions.
$P(14)=193$
Apply the stated relation and retain its conditions.
$P''=-2<0$
The feasible vertex gives the optimum; the model concerns profit, not gross revenue.

The requested value is 193.

Checks and common pitfalls: The feasible vertex gives the optimum; the model concerns profit, not gross revenue.

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

Teacher preparation and assessment

Use derivative signs, boundary values and models to justify extrema and optimization.
Which condition is essential in applications of derivatives?
Can a stationary point fail to be an extremum?

Defining relation: Use derivative signs, boundary values and models to justify extrema and optimization.
$f\prime>0\Rightarrow f\text{ increasing}$
Conditions: A critical point needs a sign or second-derivative check; closed-interval global extrema also require endpoints.