Applying calculus
Introduced| English | Meaning |
|---|---|
| optimisation/ˌɒptɪmaɪˈzeɪʃn/ | Finding the greatest or least attainable value of an objective while satisfying the stated constraints. |
| constraint/kənˈstreɪnt/ | A condition or limit that a valid solution must satisfy. |
| related rates/rɪˈleɪtɪd reɪts/ | Rates of change linked by a relationship between quantities, analysed by differentiating that relationship with respect to time. |
| area between two curves/ˈeərɪə bɪˈtwiːn tuː kɜːvz/ | area between two curves |
| marginal cost/ˈmɑːdʒɪnl kɒst/ | The additional total cost of producing one more unit of output. |
| marginal revenue/ˈmɑːdʒɪnl ˈrevənjuː/ | The additional total revenue obtained from selling one extra unit of output. |
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.
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State the target and feasible domain, use the constraint where needed, find stationary candidates, and compare classification, boundaries and any discrete restrictions.
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.
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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.
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?
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Write the chain first, then fill in each factor. It turns the question into two ordinary derivatives.
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?
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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.
In a differentiable continuous cost model, marginal cost is the derivative of total cost.
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It is an instantaneous model rate. An actual discrete one-unit increase is a finite difference and may not equal that derivative exactly.
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.
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Example: The radius is approximately 4.30 cm, minimising surface area under the stated closed-cylinder volume model.
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.