Applying calculus
| English | Chinese | Pinyin |
|---|---|---|
| optimisation | 最优化 | zuì yōu huà |
| constraint | 约束条件 | yuē shù tiáo jiàn |
| related rates | 相关变化率 | xiāng guān biàn huà lǜ |
| area between two curves | 两曲线间面积 | liǎng qū xiàn jiān miàn jī |
| marginal cost | 边际成本 | biān jì chéng běn |
| marginal revenue | 边际收益 | biān jì shōu yì |
The can that uses the least metal
- A drinks can must hold a fixed volume. Make it tall and thin, or short and wide, and the metal needed changes.
- Somewhere between is the shape that uses least. Optimisation 最优化 finds it.
- This is the unit where calculus stops being about curves and starts answering questions someone would pay for.
The optimisation method
- Write the quantity to be optimised as a function of one variable.
- Use the constraint 约束条件 to eliminate the other variable.
- Differentiate, set to zero, solve, and classify with the second derivative.
- Answer the question that was asked — often the radius, not the area.
Put the steps of an optimisation question in order.
The substitution step is where most method marks sit — it is what makes the differentiation possible.
A cylindrical can must hold 500 cm³. What radius uses least metal?
Surface area: $A = 2\pi r^2 + 2\pi rh$. Constraint: $\pi r^2 h = 500$, so $h = \dfrac{500}{\pi r^2}$.
The substitution line is where most of the method marks are: it is what turns two variables into one, and without it the differentiation cannot even start.
A cylinder must hold 500 cm³ with least surface area. Give the radius in cm, to 2 decimal places.
r³ = 250/π gives r = 4.30 cm. Substituting the constraint first is what reduces it to one variable.
Connected rates of change
- When two quantities both change with time, their rates connect through the chain rule: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$.
- Identify what you are given and what you want, then find the link between them.
- Related rates 相关变化率 questions are almost always solved by writing that chain first and filling it in afterwards.
A balloon's radius grows at a known rate and you want how fast its volume grows. What do you write first?
Write the chain first, then fill in each factor. It turns the question into two ordinary derivatives.
Area between curves, and economics
- The area between two curves 两曲线间面积 is $\displaystyle\int_a^b (f(x) - g(x))\,dx$, with $f$ the upper curve.
- Find the intersection points first — they are the limits.
- Marginal cost 边际成本 and marginal revenue 边际收益 are derivatives of total cost and total revenue, which is why economics departments credit this module.
- Profit is maximised where marginal revenue equals marginal cost, and that is a stationary-point question in disguise.
To find the area between two curves, what must you find before integrating?
The intersections are the limits of the integral, and the upper curve goes first in the subtraction.
Marginal cost is the derivative of total cost.
It is the cost of one more unit, which is a rate of change — and why economics credits this module.
Every applied answer needs units and a sentence. "$r = 4.30$" is incomplete; "the radius should be 4.30 cm, which uses about 349 cm² of metal" answers the question. At this level the interpretation is marked separately from the calculation.
You found r = 4.30 for the can. Write the full answer sentence a marker wants.
Example: "The can should have a radius of 4.30 cm, which minimises the metal used."
Check whether your stationary point is the maximum or the minimum the question wanted. An optimisation answer that finds the shape using the most metal is arithmetically perfect and worth almost nothing, and the second derivative is what tells you which one you found.