Optimisation with constraints, boundaries and rejected roots
도입됨| English | 의미 |
|---|---|
| feasible domain/ˈfiːzɪbl dəˈmeɪn/ | 가능 영역 |
Cutting larger corner squares makes a box taller but its base smaller. Which cut actually gives the largest valid volume?
- Cutting larger corner squares makes a box taller but its base smaller. Which cut actually gives the largest valid volume?
- This lesson studies feasible domain 可行域: The allowed input values after the original physical or mathematical constraints are applied.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
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.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
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A′=12−2x=0 gives x=6 in the domain.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every solution of V′=0 is a feasible global maximum of the box volume. This claim is false. Explain which definition or assumption it violates.
Find its maximum area in cm².
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A(6)=6×6=36; the derivative changes + to −.
Every solution of V′=0 is a feasible global maximum of the box volume.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the maximum of (x−1)² on [0,3].
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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.
Use this in your course
- 7357 · A-level · G. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The allowed input values after the original physical or mathematical constraints are applied. Choose the relationship, show the method, check its assumptions and interpret the result.