Applying calculus · 微积分的应用
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| optimisation/ˌɒptɪmaɪˈzeɪʃn/ | 最优化 | zuì yōu huà |
| constraint/kənˈstreɪnt/ | 约束条件 | yuē shù tiáo jiàn |
| related rates/rɪˈleɪtɪd reɪts/ | 相关变化率 | xiāng guān biàn huà lǜ |
| area between two curves/ˈeərɪə bɪˈtwiːn tuː kɜːvz/ | 两曲线间面积 | liǎng qū xiàn jiān miàn jī |
| marginal cost/ˈmɑːdʒɪnl kɒst/ | 边际成本 | biān jì chéng běn |
| marginal revenue/ˈmɑːdʒɪnl ˈrevənjuː/ | 边际收益 | biān jì shōu yì |
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 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 $A(r)=2\pi r^2+1000/r$, 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. · 把最优化题的步骤排序。
State the target and feasible domain, use the constraint where needed, find stationary candidates, and compare classification, boundaries and any discrete restrictions. · 陈述目标函数和可行定义域，必要时使用约束条件，找到驻点候选值，并比较分类、边界及任何离散限制。
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. · 一个带平顶和平底的封闭圆柱体必须容纳 500 cm³。忽略接缝，求使表面积最小化的正半径（精确到 2 位小数）。
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（r > 0）。其导数得出r³ = 250/π，故r ≈ 4.30 cm。正二阶导数及端点面积的发散性支持全局最小值的成立。
Connect rates through a defined dependence
- When $V=V(r)$ and $r=r(t)$, 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? · 气球半径以已知速率增大,要求体积增长有多快。第一步应该写什么?
Write the chain first, then fill in each factor. It turns the question into two ordinary derivatives. · 先写出这条链,再逐项填入。它把问题化为两个普通的导数。
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 边际成本 and 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? · 对于由两条相交曲线围成的有限区域，为确定积分上下限必须求出什么？
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. · 找出相关交点，确定每个区间上的上方曲线，并在顺序改变时分段。若问题指定外部区间，则可直接提供上下限。
In a differentiable continuous cost model, marginal cost is the derivative of total cost. · 在可微分的连续成本模型中，边际成本是总成本的导数。
It is an instantaneous model rate. An actual discrete one-unit increase is a finite difference and may not equal that derivative exactly. · 它是一个瞬时模型速率。实际的离散单位增量是一个有限差分，可能不等于该导数的精确值。
A derivative is not always an exact one-unit increase. For $C(q)=100+3q+0.2q^2$, $C'(10)=7$ 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. · 你算出罐子的 r = 4.30。写出阅卷人想看到的完整作答句。
Example: The radius is approximately 4.30 cm, minimising surface area under the stated closed-cylinder volume model. · 示例：在给定封闭圆柱体体积模型下，半径约为⟦4.30⟧ cm，可使表面积最小化。
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.