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Optimization Modeling and Solving in IB Mathematics | IB数学：最优化问题的建模与求解

2026-08-30 15:06 · 本文部分内容由AI辅助创作

📚 Optimization Modeling and Solving in IB Mathematics | IB数学：最优化问题的建模与求解

Optimization is one of the most practical applications of calculus in the IB Mathematics curriculum. It asks a simple but powerful question: among all possible choices, which one gives the best outcome?

最优化是IB数学课程中微积分最实用的应用之一。它提出了一个简单却有力的问题：在所有可能的选择中，哪一种能带来最佳结果？

1. What is an Optimization Problem? | 什么是最优化问题？

An optimization problem involves finding the maximum or minimum value of a function under given conditions. In IB Mathematics AA and AI, these problems often come from geometry, economics, physics, or engineering.

最优化问题是指在给定条件下求函数的最大值或最小值。在IB数学分析与方法（AA）和应用与解释（AI）中，这些问题通常来自几何、经济、物理或工程。

The key components are a quantity to be optimized and a set of rules that limit the possible solutions. For example, a farmer wishes to enclose the largest possible area with a fixed length of fencing.

关键要素包括一个待优化的量，以及一组限制可能解的条件。例如，农民希望用固定长度的篱笆围出尽可能大的面积。

2. The Modeling Process | 建模过程

Step 1: Define all relevant variables clearly and state their units.

步骤1：清晰地定义所有相关变量，并说明其单位。

Step 2: Express the quantity to be maximized or minimized as a function of these variables.

步骤2：将需要最大化或最小化的量表示成这些变量的函数。

Step 3: Use constraints to eliminate variables until only one independent variable remains, or keep several variables for multivariable cases.

步骤3：利用约束消去变量，直到只剩下一个自变量，或者多变量情形下保留多个变量。

Step 4: Apply calculus or another appropriate technique to find candidate solutions.

步骤4：运用微积分或其他适当技巧寻找候选解。

Step 5: Interpret the answer in the original context and verify its reasonableness.

步骤5：将答案放回原始情境中解释，并检验其合理性。

3. Objective Functions and Constraints | 目标函数与约束

An objective function describes the quantity we want to optimize. A constraint is an equation or inequality that the variables must satisfy.

目标函数描述了我们想要优化的量。约束是变量必须满足的方程或不等式。

For instance, suppose we want to build a rectangular box with a square base and a fixed volume V. Let x be the side of the base and h be the height. The constraint is:

例如，我们要制作一个底面为正方形、体积固定为 V 的长方体盒子。设底面边长为 x，高为 h。约束为：

x²h = V

mize the surface area S = 2x² + 4xh. Using the constraint, h = V/x², so S(x) = 2x² + 4V/x.

我们想要最小化表面积 S = 2x² + 4xh。利用约束 h = V/x²，得到 S(x) = 2x² + 4V/x。

4. Single-Variable Optimization | 单变量优化

al maxima and minima of a function f(x). The first step is to find critical points where f'(x) = 0 or where f'(x) does not exist.

在单变量最优化中，我们寻找函数 f(x) 的局部最大值和最小值。第一步是找出临界点，即 f'(x)=0 或 f'(x) 不存在的点。

lems, the function is usually differentiable on an open interval, so we focus on solving f'(x)=0.

在IB题目中，函数通常在开区间内可导，因此我们主要求解 f'(x)=0。

For example, for f(x) = x³ - 3x, we have f'(x) = 3x² - 3 = 0, which gives x = ±1.

例如，对于 f(x) = x³ - 3x，由 f'(x)=3x²-3=0 得 x=±1。

5. The Derivative Tests | 导数判别法

To classify a critical point, we can use the first derivative test or the second derivative test.

为了判断临界点的类型，我们可以使用一阶导数判别法或二阶导数判别法。

First derivative test: Examine the sign of f'(x) on either side of the critical point. If it changes from positive to negative, the point is a local maximum; if it changes from negative to positive, it is a local minimum.

一阶导数判别法：考察临界点两侧 f'(x) 的符号。若由正变负，则为局部极大值；由负变正，则为局部极小值。

Second derivative test: Compute f''(x). If f''(a)>0, then f has a local minimum at a; if f''(a)<0, then f has a local maximum. If f''(a)=0, the test is inconclusive.

二阶导数判别法：计算 f''(x)。若 f''(a)>0，则 f 在 a 处取得局部极小值；若 f''(a)<0，则取得局部极大值。若 f''(a)=0，则判别法失效。

6. Optimizing on a Closed Interval | 闭区间上的最优化

When the domain is a closed interval [a,b], the global maximum and minimum must occur either at critical points inside the interval or at the endpoints a and b.

当定义域是闭区间 [a,b] 时，全局最大值和最小值必然出现在区间内的临界点或端点 a、b 处。

The procedure is: find all critical points in (a,b), evaluate f at those points and at the endpoints, and compare the values.

步骤为：找出 (a,b) 内所有临界点，计算这些点以及端点处的函数值，然后比较大小。

in IB Paper 2, especially when a real-world context limits the domain, for example x ≥ 0.

这在IB Paper 2中经常考查，尤其是当实际情境限制了定义域，例如 x≥0。

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7. Multi-Variable Optimization | 多变量优化

ariables, z = f(x,y). Critical points occur where both first partial derivatives are zero:

在IB数学AA HL中，学生将最优化推广到二元函数 z=f(x,y)。临界点出现在两个一阶偏导数均为零的地方：

∂f/∂x = 0 and ∂f/∂y = 0

minimum; if D>0 and ∂²f/∂x²<0, there is a local maximum; if D<0, there is a saddle point.

二元函数的二阶导数判别法使用判别式 D=(∂²f/∂x²)(∂²f/∂y²)-(∂²f/∂x∂y)²。若 D>0 且 ∂²f/∂x²>0，则为局部极小值；若 D>0 且 ∂²f/∂x²<0，则为局部极大值；若 D<0，则为鞍点。

8. Constrained Optimization and Lagrange Multipliers | 约束优化与拉格朗日乘子

When optimizing a function of several variables subject to an equality constraint, the method of Lagrange multipliers is powerful.

在等式约束下优化多元函数时，拉格朗日乘子法是一种强有力的工具。

To maximize or minimize f(x,y) subject to g(x,y)=k, solve the system ∇f = λ∇g together with g(x,y)=k.

在约束 g(x,y)=k 下最大化或最小化 f(x,y)，需要求解方程组 ∇f = λ∇g 以及 g(x,y)=k。

Here λ is the Lagrange multiplier. This method appears in IB Mathematics AA HL only.

其中 λ 为拉格朗日乘子。此方法仅在IB数学AA HL中涉及。

we set 2x=λ, 2y=λ, and x+y=1. This gives x=y=1/2, corresponding to the minimum value 1/2.

例如，在 x+y=1 约束下求 f(x,y)=x²+y² 的最值，令 2x=λ，2y=λ，x+y=1。解得 x=y=1/2，对应最小值为 1/2。

9. Worked Example: Maximizing Volume | 实例：最大化体积

Problem: A rectangular box with a square base and no lid is made from 12 m² of cardboard. Find the maximum volume.

问题：用 12 m² 的硬纸板制作一个无盖长方体盒子，底面为正方形，求最大体积。

nd h be the height. The surface area constraint is x² + 4xh = 12, because there is no lid.

设底面边长为 x，高为 h。由于无盖，表面积约束为 x²+4xh=12。

Then h = (12 - x²)/(4x). The volume is V = x²h = x(12 - x²)/4 = 3x - x³/4.

于是 h=(12-x²)/(4x)。体积为 V=x²h=x(12-x²)/4=3x - x³/4。

etting this to zero gives x=2, since x>0. Then h=(12-4)/8=1. The maximum volume is V=4 m³.

求导得 dV/dx=3-3x²/4。令其为零得 x=2（因为 x>0）。则 h=(12-4)/8=1。最大体积为 V=4 m³。

The second derivative is d²V/dx² = -3x/2, which is negative for x>0, confirming a maximum.

二阶导数为 d²V/dx²=-3x/2，在 x>0 时为负，确认为极大值。

10. Worked Example: Minimizing Cost | 实例：最小化成本

Problem: A cylindrical can with a fixed volume of 16π cm³ must have the minimum total surface area. Find the radius and height.

问题：一个体积固定为 16π cm³ 的圆柱形罐头，要求总表面积最小，求底面半径和高。

The volume constraint gives πr²h = 16π, so h = 16/r².

体积约束给出 πr²h=16π，因此 h=16/r²。

The surface area is S = 2πr² + 2πrh = 2πr² + 32π/r.

表面积为 S=2πr²+2πrh=2πr²+32π/r。

ing with respect to r gives dS/dr = 4πr - 32π/r² = 0, so r³ = 8, hence r=2. Then h=16/4=4.

对 r 求导得 dS/dr=4πr-32π/r²=0，因此 r³=8，即 r=2。于是 h=16/4=4。

Since d²S/dr² = 4π + 64π/r³ > 0 for all r>0, this critical point gives a minimum.

由于 d²S/dr²=4π+64π/r³ 对所有 r>0 均大于0，所以该临界点对应极小值。

11. Common Pitfalls | 常见误区