Applying calculus · Aplicar cálculo
| English | Español |
|---|---|
| optimisation/ˌɒptɪmaɪˈzeɪʃn/ | optimización |
| constraint/kənˈstreɪnt/ | restricción |
| related rates/rɪˈleɪtɪd reɪts/ | tasas relacionadas |
| area between two curves/ˈeərɪə bɪˈtwiːn tuː kɜːvz/ | área entre dos curvas |
| marginal cost/ˈmɑːdʒɪnl kɒst/ | el costo marginal |
| marginal revenue/ˈmɑːdʒɪnl ˈrevənjuː/ | ingreso marginal |
A fixed volume constrains the design
- A closed cylindrical teaching model holds 500 cubic centimetres, with flat top and bottom and no allowance for seams. Its radius and height cannot be chosen independently.
- Optimisation · Optimización 最优化 finds a required maximum or minimum over a stated feasible domain; the physical assumptions are part of the model.
Reduce variables and compare candidates
- Use the constraint 约束条件 to express the target as one variable. For a closed cylinder, $h=500/(\pi r^2)$ and · y $A(r)=2\pi r^2+1000/r$, with $r>0$.
- Differentiate to find stationary candidates, then justify classification and check applicable boundaries. For integer choices, compare feasible neighbouring integers rather than treating a fractional optimum as an allowed answer.
Put the steps of an optimisation question in order. · Ordena los pasos de una pregunta de optimización.
State the target and feasible domain, use the constraint where needed, find stationary candidates, and compare classification, boundaries and any discrete restrictions. · Establece el dominio objetivo y factible, utiliza la restricción cuando sea necesario, encuentra candidatos estacionarios y compara clasificación, fronteras y cualquier restricción discreta.
A closed-cylinder minimum. $A'(r)=4\pi r-1000/r^2=0$ gives $r^3=250/\pi$, hence $r\approx4.30$ cm. Since $A''(r)=4\pi+2000/r^3>0$ for positive r and A grows without bound at either end, this is the global minimum of the stated continuous model. The volume constraint then gives $h=2r$.
A closed cylinder with flat top and bottom must hold 500 cm³. Ignoring seams, find the radius minimising surface area for positive radius, to 2 decimal places. · Un cilindro cerrado con tapa y base planas debe contener 500 cm³. Ignorando las costuras, halla el radio que minimice el área superficial para un radio positivo, con 2 decimales.
A(r) = 2πr² + 1000/r for r > 0. Its derivative gives r³ = 250/π, so r ≈ 4.30 cm. The positive second derivative and divergent endpoint areas justify the global minimum. · A(r) = 2πr² + 1000/r para r > 0. Su derivada da r³ = 250/π, por lo que r ≈ 4.30 cm. La segunda derivada positiva y el divergencia de las áreas en los extremos justifican el mínimo global.
Connect rates through a defined dependence
- When $V=V(r)$ and · y $r=r(t)$, the chain rule gives $dV/dt=(dV/dr)(dr/dt)$.
- Related rates 相关变化率 need compatible units and values at the requested instant. Integrating a rate gives accumulated change, while an initial amount determines the final total.
A balloon's radius grows at a known rate and you want how fast its volume grows. What do you write first? · El radio de un globo crece a una tasa conocida y quieres saber qué tan rápido crece su volumen. ¿Qué escribes primero?
Write the chain first, then fill in each factor. It turns the question into two ordinary derivatives. · Escribe primero la regla de la cadena, luego rellena cada factor. Convierte la pregunta en dos derivadas ordinarias.
Areas and marginal models have limits
- Area between two curves 两曲线间面积 uses upper minus lower over the specified region. Find intersections for enclosed regions and split at order changes; an externally specified interval may instead supply the limits.
- Marginal cost · Costo marginal 边际成本 and · y marginal revenue 边际收益 are derivatives of continuous models. Equating them finds an interior stationary profit candidate, whose feasibility, classification and boundary competitors still need checking.
For the finite region enclosed by two intersecting curves, what must you find to determine its integration bounds? · Para la región finita encerrada entre dos curvas que se intersectan, ¿qué debes encontrar para determinar sus límites de integración?
Find the relevant intersections, determine which curve is upper on each interval, and split if their order changes. A question specifying an external interval can instead supply the bounds directly. · Encuentra las intersecciones relevantes, determina qué curva está arriba en cada intervalo y divide si su orden cambia. Una pregunta que especifique un intervalo externo puede proporcionar directamente los límites.
In a differentiable continuous cost model, marginal cost is the derivative of total cost. · En un modelo de costos continuo diferenciable, el costo marginal es la derivada del costo total.
It is an instantaneous model rate. An actual discrete one-unit increase is a finite difference and may not equal that derivative exactly. · Es una tasa de modelo instantáneo. Un aumento real discreto de una unidad es una diferencia finita y puede no ser exactamente igual a esa derivada.
A derivative is not always an exact one-unit increase. For · A favor $C(q)=100+3q+0.2q^2$, $C'(10)=7$ but $C(11)-C(10)=7.2$. Sheet 3.3 also compares a continuous maximum at 5.5 with the best whole-number quantities 5 and 6.
You found r = 4.30 for the can. Write the full answer sentence a marker wants. · Encontraste r = 4.30 para la lata. Escribe la oración completa de respuesta que busca un evaluador.
Example: The radius is approximately 4.30 cm, minimising surface area under the stated closed-cylinder volume model. · Ejemplo: El radio es aproximadamente 4.30 cm, minimizando el área superficial bajo el modelo de volumen de cilindro cerrado establecido.
A stationary point alone does not establish the global optimum. State the feasible domain and why other candidates or boundaries cannot give a better value. A different container opening, seam allowance or practical dimension constraint can change the model.