Optimization in IB Mathematics | IB数学中的最优化

📚 Optimization in IB Mathematics | IB数学中的最优化

In the real world, we often seek the best outcome: the highest profit, the shortest time, or the minimum material used. In IB Mathematics, optimization is the mathematical tool that turns these questions into solvable problems. It primarily relies on differential calculus to find maximum and minimum values of functions derived from real-life situations. Whether you are in the Analysis and Approaches or Applications and Interpretation course, mastering optimization is essential for both internal assessments and final examinations.

在现实世界中，我们常常追求最佳结果：最高利润、最短时间或最小材料用量。在IB数学中，最优化就是将这些实际问题转化为可解问题的数学工具。它主要依靠微分学来寻找从实际情景中导出的函数的最大值和最小值。无论你选择的是分析与方法还是应用与解释课程，掌握最优化对于内部评估和最终考试都至关重要。

1. What is Optimization? | 什么是优化？

Optimization is the process of finding the best solution from all feasible solutions. In mathematical terms, it means determining the maximum or minimum value of a function, called the objective function, under given constraints. For example, a company may want to maximise profit, or an engineer may design a can with minimum surface area for a fixed volume. In IB Mathematics, optimisation problems typically involve forming a single-variable function and then using calculus to locate its extreme values.

优化是从所有可行方案中寻找最佳方案的过程。数学上讲，就是在给定约束条件下，确定一个函数（称为目标函数）的最大值或最小值。例如，公司可能希望最大化利润，或者工程师可能设计一个给定容积下表面积最小的易拉罐。在IB数学中，最优化问题通常需要建立一个单变量函数，然后利用微积分找到它的极值。

The word ‘optimum’ can refer to a global (absolute) maximum or minimum over the entire domain, or a local (relative) extremum within a small interval. IB problems usually ask for the ‘maximum’ or ‘minimum’ in context, so you must interpret the result carefully.

“最优”可以指整个定义域上的全局（绝对）最大值或最小值，也可以指小区间内的局部（相对）极值。IB题目通常要求在实际背景下求“最大”或“最小”，因此你必须仔细解释结果。

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2. The General Procedure for Solving Optimization Problems | 求解优化问题的一般步骤

The first step is to read the problem carefully and determine what needs to be maximised or minimised, such as area, volume, cost, or profit. Give this quantity a name, for example, P for profit or V for volume.

第一步是仔细阅读问题，确定需要最大化或最小化的量，比如面积、体积、成本或利润。给这个量起一个名字，例如用 P 表示利润，用 V 表示体积。

Next, introduce variables to represent the unknowns, and express the quantity to be optimised as a function of one variable using any given constraints. This often involves substituting a relationship to eliminate extra variables.

接下来，引入变量表示未知数，并利用给定的约束条件将待优化的量表示为一个变量的函数。这一步通常需要用已知关系进行代入，以消去多余的变量。

Determine the feasible domain of the independent variable. For geometric problems, this is often an interval determined by physical limitations, such as positive lengths. Check whether endpoints should be considered for absolute extrema.

确定自变量的可行定义域。对于几何问题，这通常是由物理限制（如长度为正）确定的一个区间。需要考虑是否要检查端点处的绝对极值。

Find the first derivative of the objective function with respect to the chosen variable. Set the derivative equal to zero and solve for the variable to obtain critical points. These are candidates for local extrema.

求目标函数关于所选变量的一阶导数。令导数等于零并求解变量，得到临界点。这些点是局部极值的候选点。

Use the second derivative test or analyse the sign change of the first derivative to classify each critical point as a local maximum or minimum. If the domain is a closed interval, evaluate the function at the critical points and at the endpoints to identify the absolute maximum or minimum.

使用二阶导数检验，或分析一阶导数的符号变化，来判断每个临界点是局部最大值还是最小值。如果定义域是闭区间，还要计算函数在临界点和端点处的值，以确定绝对最大值或最小值。

Finally, interpret the mathematical result in the context of the original problem. State the optimum value and the corresponding values of all variables with appropriate units.

最后，将数学结果解释回原问题背景。说明最优值以及所有变量的相应取值，并带上合适的单位。

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3. Identifying the Objective Function and Constraints | 识别目标函数与约束条件

The objective function is the expression you want to optimise, such as A = xy for the area of a rectangle. A constraint is an equation linking the variables, for example, 2x + 2y = 100 for a fixed perimeter. Your task is to use the constraint to rewrite the objective function in terms of a single variable.

目标函数是你想要优化的表达式，例如矩形的面积 A = xy。约束条件是联系变量的方程，例如周长固定时 2x + 2y = 100。你的任务就是利用约束条件将目标函数重新表示为仅含一个变量的函数。

In many IB optimization problems, the constraint is not explicitly given as an equation; you need to derive it from the scenario. For instance, if a box is made from a flat sheet by cutting squares from the corners, the length, width, and height are related. Identifying these relationships correctly is critical.

在许多IB最优化问题中，约束条件并非以方程的形式直接给出；你需要从情景中推导出来。例如，如果用一块平板通过切去四角来制作盒子，那么长、宽和高之间就存在关联。正确识别这些关系至关重要。

Always assign clear variable names and draw a diagram if possible. A well-labelled sketch can make the constraint visible and help you avoid sign errors or dimension mistakes.

始终为变量取清晰的名字，并尽可能画出示意图。一个标注清楚的草图可以让约束关系一目了然，帮助你避免符号错误或维度错误。

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4. Using the First Derivative to Find Critical Points | 使用一阶导数寻找临界点

Once the objective function f(x) is established, compute its derivative f'(x). The critical points occur where f'(x) = 0 or where f'(x) is undefined. In IB questions, the derivative is usually set to zero to solve for stationary points.

一旦建立了目标函数 f(x)，就计算它的导数 f'(x)。临界点出现在 f'(x) = 0 或 f'(x) 无定义的地方。在IB考题中，通常令导数等于零来求解驻点。

For a quadratic function such as f(x) = 100x – 2x², the derivative is f'(x) = 100 – 4x. Setting this equal to zero gives x = 25. This is the only critical point in the domain.

对于像 f(x) = 100x – 2x² 这样的二次函数，导数为 f'(x) = 100 – 4x。令其等于零得到 x = 25。这就是定义域内唯一的临界点。

Be careful when the derivative leads to a rational equation or requires factorisation. Always simplify the derivative to a form that can be solved algebraically. In some IB applications, you may be allowed to use a GDC to solve f'(x) = 0, but you must still show the derivative set-up.

当导数是一个有理方程或需要进行因式分解时要格外小心。始终将导数化简为可以代数求解的形式。在某些IB应用中，你可以使用图形计算器求解 f'(x) = 0，但仍必须展示导数的建立过程。

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5. Second Derivative Test for Local Extrema | 二阶导数检验局部极值

The second derivative test provides a quick way to classify a stationary point. If f”(c) > 0, the function has a local minimum at x = c. If f”(c) < 0, the function has a local maximum. If f''(c) = 0, the test is inconclusive and you must use the first derivative test.

二阶导数检验提供了一种快速判断驻点类型的方法。如果 f”(c) > 0，则函数在 x = c 处有局部极小值；如果 f”(c) < 0，则有局部极大值。如果 f''(c) = 0，该检验无法得出结论，你需要使用一阶导数检验。

Sign of f”(c)	Type of Critical Point
f”(c) > 0	Local minimum (concave up)
f”(c) < 0	Local maximum (concave down)
f”(c) = 0	Test fails; use first derivative test

This table summarises how the concavity given by the second derivative determines the nature of the stationary point. In IB exams, showing both f'(x) and f”(x) strengthens your reasoning and avoids ambiguous conclusions.

这个表格总结了二阶导数给出的凹凸性如何决定驻点的性质。在IB考试中，同时展示 f'(x) 和 f”(x) 可以加强你的推理，避免模棱两可的结论。

For the example f(x) = 100x – 2x², we have f'(x) = 100 – 4x and f”(x) = -4. Since f”(25) = -4 < 0, the point is a local maximum, which confirms the largest area.

对于例子 f(x) = 100x – 2x²，我们有 f'(x) = 100 – 4x，f”(x) = -4。因为 f”(25) = -4 < 0，该点是一个局部极大值，这确认了最大面积。

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6. Closed Interval Method for Absolute Extrema | 闭区间方法求绝对极值

When the domain of the function is restricted to a closed interval [a, b], a local extremum is not necessarily the absolute extremum. You must evaluate the function at all critical points inside the interval and at the endpoints a and b. The largest of these values is the absolute maximum, and the smallest is the absolute minimum.

当函数的定义域被限定在一个闭区间 [a, b] 上时，局部极值未必是绝对极值。你必须计算函数在区间内所有临界点以及端点 a 和 b 处的值。这些值中的最大者就是绝对最大值，最小者就是绝对最小值。

This method is especially relevant in problems where the variable represents a physical quantity with natural bounds. For instance, a length cannot be negative, and it also cannot exceed a certain value determined by the available material. Evaluating at these endpoints may yield a higher or lower value than the stationary point.

当变量代表具有天然界限的物理量时，这种方法尤为重要。例如，长度不能为负，且不能超过由可用材料决定的某个值。在端点处求值可能得到比驻点更高或更低的结果。

In IB mark schemes, final answers that do not check endpoints on a closed domain may lose a mark if an endpoint gives the optimum. Always define the domain explicitly and mention whether the endpoints have been considered.

在IB评分标准中，如果在闭区间上没有检查端点，而恰好在端点处取得最优值，可能会被扣分。务必明确定义域，并说明是否已考虑端点。

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7. Geometric Optimization: Minimising Surface Area | 几何优化：最小化表面积

A classic problem is to design a cylindrical can that holds a fixed volume V while using the minimum amount of metal. Let the radius be r and the height be h. The volume constraint is V = πr²h, and the surface area is S = 2πr² + 2πrh. Express h in terms of r from the constraint: h = V/(πr²).

一个经典问题是设计一个圆柱形易拉罐，使其在容纳固定体积 V 的同时使用最少的金属材料。设半径为 r，高为 h。体积约束为 V = πr²h，表面积为 S = 2πr² + 2πrh。从约束条件中把 h 用 r 表示：h = V/(πr²)。

Substitute into S: S(r) = 2πr² + 2πr × V/(πr²) = 2πr² + 2V/r. Differentiate: S'(r) = 4πr – 2V/r². Set S'(r) = 0: 4πr = 2V/r² → 4πr³ = 2V → r³ = V/(2π), so r = (V/(2π))^(1/3). The second derivative S”(r) = 4π + 4V/r³ is positive for all positive r, confirming a minimum.

代入 S：S(r) = 2πr² + 2πr × V/(πr²) = 2πr² + 2V/r。求导：S'(r) = 4πr – 2V/r²。令 S'(r) = 0：4πr = 2V/r² → 4πr³ = 2V → r³ = V/(2π)，因此 r = (V/(2π))^(1/3)。二阶导数 S”(r) = 4π + 4V/r³ 对所有正数 r 都为正，确认了这是最小值。

This problem beautifully demonstrates the power of calculus to solve practical design questions. IB examinations frequently include such geometric optimisation tasks, requiring clear setting out, correct differentiation, and a concluding statement with units.

这个问题很好地展示了微积分解决实际设计问题的能力。IB考试中经常出现这类几何最优化任务，要求解题过程清晰、求导正确，并给出带单位的结论。

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8. Economic Optimization: Maximising Profit | 经济优化：最大化利润

In economics, profit P(x) is usually the difference between revenue R(x) and cost C(x), where x is the number of units produced and sold. The objective is to maximise P(x) = R(x) – C(x). The first derivative P'(x) = R'(x) – C'(x) gives the marginal profit. Critical points occur when marginal revenue equals marginal cost.

在经济学中，利润 P(x) 通常是收入 R(x) 与成本 C(x) 之差，其中 x 是生产和销售的单位数量。目标是最大化 P(x) = R(x) – C(x)。一阶导数 P'(x) = R'(x) – C'(x) 给出边际利润。临界点出现在边际收入等于边际成本时。

A typical IB problem provides a demand function p(x) (price per unit), from which revenue R(x) = x·p(x). Cost might be a quadratic or linear function. Students must build the profit function, find its maximum, and also confirm that it is indeed a maximum using the second derivative.

一个典型的IB题目会给出需求函数 p(x)（单位价格），从而收入 R(x) = x·p(x)。成本可能是二次函数或线性函数。学生需要建立利润函数，求出最大值，并用二阶导数确认它确实是最大值。

For example, if p = 200 – 2x and C = 1000 + 20x, then R = x(200 – 2x) = 200x – 2x², so P = 200x – 2x² – (1000 + 20x) = -2x² + 180x – 1000. Then P'(x) = -4x + 180 = 0 gives x = 45. P”(x) = -4 < 0, so maximum profit occurs at 45 units. Always note the feasible domain, often x ≥ 0 and p ≥ 0.

例如，若 p = 200 – 2x，C = 1000 + 20x，则 R = x(200 – 2x) = 200x – 2x²，于是 P = 200x – 2x² – (1000 + 20x) = -2x² + 180x – 1000。然后 P'(x) = -4x + 180 = 0，得 x = 45。P”(x) = -4 < 0，所以最大利润出现在 45 个单位时。务必注意可行定义域，通常 x ≥ 0 且 p ≥ 0。

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9. Optimization in Kinematics: Minimising Distance | 运动学优化：最小化距离

Optimization also appears in kinematics when we want to minimise the distance between two moving objects or find the time at which velocity is minimised. For a position function s(t), the velocity v(t) = s'(t). To minimise velocity magnitude, we can differentiate the velocity function and set the derivative to zero.

最优化也出现在运动学中，当我们需要最小化两个运动物体间的距离，或者找到速度最小时的时间。对于位置函数 s(t)，速度 v(t) = s'(t)。要最小化速度的大小，我们可以对速度函数求导并令导数为零。

A typical problem: given s(t) = t³ – 9t² + 24t, find the minimum speed over the first 5 seconds. First find v(t) = 3t² – 18t + 24. The acceleration a(t) = v'(t) = 6t – 18. Setting a(t) = 0 gives t = 3. Check v(0), v(3), v(5) and take the smallest magnitude. Here v(3) = 3(9) – 18(3) + 24 = 27 – 54 + 24 = -3, speed 3. While at t=0 speed is 24, at t=5 speed is positive relatively larger. Minimum speed is 3 units.

一个典型问题：已知 s(t) = t³ – 9t² + 24t，求在最初5秒内的最小速率。首先求 v(t) = 3t² – 18t + 24。加速度 a(t) = v'(t) = 6t – 18。令 a(t) = 0 得 t = 3。检查 v(0)、v(3)、v(5) 并取绝对值最小的。这里 v(3) = 3(9) – 18(3) + 24 = 27 –

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