Optimization Problem Solving Strategies | 优化问题求解策略

📚 Optimization Problem Solving Strategies | 优化问题求解策略

Optimization is one of the most practical applications of calculus in the IB Mathematics curriculum. It involves finding the maximum or minimum value of a function under given constraints, often in real-world contexts.

优化问题是 IB 数学课程中微积分最具实际应用价值的内容之一。它涉及在给定约束条件下寻找函数的最大值或最小值，通常应用于现实情境中。

1. Understanding Optimization Problems | 理解优化问题

An optimization problem typically asks you to find the largest or smallest possible value of a quantity. This quantity is modeled by a function, and the domain of the function is often restricted by physical or geometric constraints.

优化问题通常要求你找出某个量的最大或最小值。这个量由一个函数建模，而函数的定义域往往受到物理或几何条件的限制。

Identify the quantity to be optimized (e.g., area, volume, cost, profit).

确定要优化的量（如面积、体积、成本、利润）。
Identify all constraints given in the problem.

找出题目中的所有约束条件。
Express the target quantity as a function of one variable.

将目标量表示为一个变量的函数。

2. Key Steps: The Optimization Framework | 关键步骤：优化框架

Solving an optimization problem follows a systematic approach. Use this framework to organize your work in the exam.

解决优化问题需要遵循系统化的步骤。在考试中使用以下框架来组织你的解题过程。

Draw a diagram if possible and label variables.

如果可能，画出示意图并标注变量。
Write down the quantity to be optimized.

写出需要优化的量。
Use constraints to reduce the function to one variable.

利用约束条件将函数化为单变量函数。
Find the derivative and critical points.

求导数并找出临界点。
Determine whether each critical point is a maximum or minimum.

判断每个临界点是最大值还是最小值。
Check endpoints of the domain if they exist.

如果定义域端点存在，检查端点的函数值。
State the final answer clearly with units.

清晰给出最终答案，并注明单位。

3. Example 1: Maximizing Area with a Fixed Perimeter | 例1：固定周长下最大化面积

A farmer has 100 metres of fencing and wants to enclose a rectangular field. What dimensions maximize the area?

一位农民有100米围栏，想围成一个矩形田地。什么尺寸能使面积最大？

Let the length be x and the width be y. The perimeter constraint gives:

设长为 x，宽为 y。周长约束为：

2x + 2y = 100 → y = 50 − x

The area is A = xy = x(50 − x) = 50x − x². Differentiate:

面积 A = xy = x(50 − x) = 50x − x²。求导：

A′(x) = 50 − 2x

Setting A′(x) = 0 gives x = 25, so y = 25. The maximum area is 625 m².

令 A′(x) = 0，得 x = 25，因此 y = 25。最大面积为625平方米。

4. Example 2: Minimizing Surface Area for a Fixed Volume | 例2：固定体积下最小化表面积

A cylindrical can holds a volume of 1000 cm³. Find the radius and height that minimize the total surface area.

一个圆柱形罐子的体积为1000立方厘米。求使总表面积最小的半径和高度。

Volume constraint: πr²h = 1000, so h = 1000/(πr²).

体积约束：πr²h = 1000，因此 h = 1000/(πr²)。

Surface area includes the two circular ends and the curved side:

表面积包括两个圆形底面和侧面：

S = 2πr² + 2πrh = 2πr² + 2000/r

Differentiate with respect to r:

对 r 求导：

S′(r) = 4πr − 2000/r²

Set S′(r) = 0:

令 S′(r) = 0：

4πr = 2000/r² → r³ = 500/π → r ≈ 5.42 cm

Then h = 1000/(πr²) ≈ 10.84 cm. The minimizing height is twice the radius.

则 h = 1000/(πr²) ≈ 10.84 厘米。使表面积最小的高度是半径的两倍。

5. Using the Second Derivative | 使用二阶导数

The second derivative test helps determine the nature of a critical point. If f″(x) > 0, the point is a local minimum; if f″(x) < 0, it is a local maximum.

二阶导数检验有助于判断临界点的性质。若 f″(x) > 0，则为局部极小值；若 f″(x) < 0，则为局部极大值。

For the area example, A″(x) = −2 < 0, confirming a maximum. For the cylinder example, S″(r) = 4π + 4000/r³ > 0, confirming a minimum.

对于面积例子，A″(x) = −2 < 0，确认是最大值。对于圆柱例子，S″(r) = 4π + 4000/r³ > 0，确认是最小值。

6. Checking Endpoints and Domains | 检查端点和定义域

Many optimization problems have a restricted domain. Always evaluate the function at the endpoints of the domain, because the global maximum or minimum may occur there rather than at a critical point.

许多优化问题都有受限制的定义域。务必在定义域端点处评估函数，因为全局最大值或最小值可能出现在端点而非临界点。

For example, if x ∈ [0, 50] in the fencing problem, the endpoints give A(0) = 0 and A(50) = 0, which are minima. The maximum occurs at x = 25, inside the interval.

例如，在围栏问题中，若 x ∈ [0, 50]，端点处 A(0) = 0 和 A(50) = 0 都是最小值。最大值出现在区间内部的 x = 25 处。

7. Optimization with Implicit Differentiation | 隐函数求导的优化

Sometimes the relationship between variables is not explicitly solved. You can differentiate the constraint equation implicitly and substitute into the derivative of the target function.

有时变量之间的关系无法显式解出。你可以对约束方程进行隐式求导，并代入目标函数的导数中。

Consider a right circular cone inscribed in a sphere. The constraint is r² + h² = R², where R is the sphere radius. By implicitly differentiating or substituting, you reduce the volume function to one variable.

考虑一个内接于球体的圆锥。约束为 r² + h² = R²，其中 R 是球半径。通过隐式求导或代入，可以将体积函数化为单变量函数。

8. Word Problems: Translating Language into Math | 应用题：将语言转化为数学

IB exam questions often use real-world language. Key phrases include “at most”, “at least”, “maximum possible”, “minimum cost”, and “optimize profit”. Write down the mathematical meaning of each phrase before starting.

IB 考试题常用现实语言。关键短语包括”至多”、”至少”、”最大可能”、”最小成本”和”最优利润”。在开始前写下每个短语的数学含义。

“Sum of two numbers is 20” → x + y = 20

“两数之和为20” → x + y = 20
“Product is maximized” → find max of P = xy

“乘积最大化” → 求 P = xy 的最大值
“Cost is minimized” → find min of C(x)

“成本最小化” → 求 C(x) 的最小值

9. Common Mistakes to Avoid | 常见错误避免

Many students lose marks due to avoidable errors. Be careful with units, domain restrictions, and whether you are asked for the value of the variable or the optimized quantity itself.

许多学生因可避免的错误而失分。注意单位、定义域限制，以及题目要求的是变量的值还是优化量本身。

Forgetting the domain.

忘记定义域。
Not verifying that the critical point gives the desired type of extremum.

未验证临界点给出的是期望的极值类型。
Using the wrong formula for surface area or volume.

使用错误的表面积或体积公式。
Stopping after finding the derivative without solving f′(x) = 0.

求出导数后未解 f′(x) = 0 就停止。
Ignoring endpoint values when the domain is closed.

定义域为闭区间时忽略端点值。

10. Practice Problem: Profit Maximization | 练习题：利润最大化

A company sells x units of a product. The revenue is R(x) = 80x − x² and the cost is C(x) = 200 + 20x. Find the number of units that maximizes profit.

一家公司销售 x 件产品。收入为 R(x) = 80x − x²，成本为 C(x) = 200 + 20x。求使利润最大的销售数量。

Profit = Revenue − Cost:

利润 = 收入 − 成本：

P(x) = (80x − x²) − (200 + 20x) = 60x − x² − 200

Differentiate and set to zero:

求导并令其为零：

P′(x) = 60 − 2x = 0 → x = 30

Since P″(x) = −2 < 0, x = 30 gives a maximum profit of P(30) = 60(30) − 30² − 200 = 700.

由于 P″(x) = −2 < 0，x = 30 时利润最大，最大利润为 P(30) = 60(30) − 30² − 200 = 700。

11. Optimization with Technology (GDC) | 利用图形计算器优化

The IB exam allows a graphing display calculator (GDC). You can use the maximum/minimum feature to verify your analytic answer, which is an excellent strategy for time management.

IB 考试允许使用图形计算器（GDC）。你可以使用最大值/最小值功能验证解析答案，这是管理时间的绝佳策略。

Graph the single-variable function.

画出单变量函数图像。
Use the “max” or “min” command to locate the extremum.

使用”最大值”或”最小值”命令定位极值。
Write down the exact coordinates from the screen or store them.

记下屏幕上的精确坐标或存储它们。

Always show your analytic steps as well; the GDC result alone is not sufficient for full marks in most questions.

务必同时展示你的解析步骤；在大多数问题中，仅靠 GDC 结果无法获得满分。

12. Summary and Exam Tips | 总结与考试提示

Optimization is a high-scoring topic if you follow a routine. Practise with past paper questions and memorize the formulas for common shapes such as rectangles, cylinders, cones, and spheres.

如果按固定步骤解题，优化是一个高分考点。练习历年真题，并牢记常见图形如矩形、圆柱、圆锥和球体的公式。

In the final exam, define your variables clearly, write the constraint equation, and show that you have checked the nature of the extremum. This demonstrates full understanding to the examiner.

在最终考试中，清晰地定义变量，写出约束方程，并展示你已检验极值的性质。这会向考官展示你充分的理解。

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Optimization Problem Solving Strategies | 优化问题求解策略