Optimisation with constraints, boundaries and rejected roots
Введено| English | Значение |
|---|---|
| feasible domain/ˈfiːzɪbl dəˈmeɪn/ | область допустимых значений |
Which cut actually gives the largest valid volume?
- Cutting larger corner squares makes a box taller but its base smaller.
- This lesson studies feasible domain 可行域: The allowed input values after the original physical or mathematical constraints are applied.
The method
- Express the quantity to optimise in one variable, with its feasible domain.
- Solve a derivative-zero condition for interior candidates, reject invalid roots, and classify using derivative signs or a nonzero second derivative.
- A global maximum/minimum also needs boundary values or limiting behaviour; local classification alone does not settle a restricted-domain question.
For the open box, V′=0 gives x≈2.43 and x≈8.24. Why is 8.24 rejected outright?
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A stationary point outside the feasible domain is not a candidate. Global claims also need the boundaries: V→0 at both open ends, so the interior maximum at 2.43 is global.
A worked case
Cut squares of side x cm from a 20 cm by 12 cm sheet and fold an open box. Its dimensions are x,20−2x,12−2x, so V=240x−64x²+4x³ and 0<x<6. V′=240−128x+12x² gives roots x=(16±2√19)/3. The smaller root α≈2.4274 lies in the domain; the larger ≈8.2393 makes the short base side negative and is rejected. V′ is positive on (0,α) and negative on (α,6), while V tends to zero at both ends. Thus α gives the global maximum. A separate rectangle with perimeter 24 cm has sides x,12−x, area A=12x−x² and 0<x<12; its maximum is 36 cm² at x=6. For f=(x−1)² on [0,3], the interior stationary value is the minimum 0, but the global maximum is the endpoint value f(3)=4.
Optimisation with constraints, boundaries and rejected roots
Express the quantity to optimise in one variable, with its feasible domain
Connect derivative calculations to their original point, domain and stated rate law.
For the 24 cm perimeter rectangle, find the optimum x in cm.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
A′=12−2x=0 gives x=6 in the domain.
Where it goes wrong
- A stationary root outside the feasible domain is not an alternative design.
- A second-derivative test gives a local classification, not every global comparison.
- Open endpoints are limiting values, not permitted designs.
- For a closed interval check actual endpoint values.
- If dimensions must use a specified discrete step, compare the neighbouring allowed choices rather than reporting an unattainable continuous optimum.
Not true: Every solution of V′=0 is a feasible global maximum of the box volume. A stationary point must lie inside the feasible domain, and the endpoints must be checked for the global maximum.
Find its maximum area in cm².
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
A(6)=6×6=36; the derivative changes + to −.
A stationary point must lie inside the feasible domain, and the endpoints must be checked for the global maximum.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. A stationary point must lie inside the feasible domain, and the endpoints must be checked for the global maximum.
Applying it
- Write units and constraints before calculus, keep exact roots through classification, and give the requested final dimensions or quantity rather than only the input.
- State idealisations such as negligible material thickness and exact folds.
- If the model has discrete cuts or measurement tolerance, explain how the final feasible choice and its comparison would change.
Find the maximum of (x−1)² on [0,3].
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Compare the stationary value 0 with endpoint values 1 and 4; the maximum is 4.
Match each optimisation step to its purpose.
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A second-derivative test classifies locally only; restricted-domain optimisation is finished by comparing boundaries.
In your course
- AQA A-level Mathematics (7357), specification section G.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Feasible domain: The allowed input values after the original physical or mathematical constraints are applied.
- Express the quantity to optimise in one variable, with its feasible domain.
- Watch for: A stationary root outside the feasible domain is not an alternative design.