📚 16B Optimization | 第16B节：优化
Optimization is a fundamental application of calculus that allows us to find the best possible outcome, such as maximum profit or minimum cost. In IB Mathematics, optimization problems often appear in Paper 2, requiring students to model real-world scenarios and apply differentiation techniques. This article provides a comprehensive guide to solving optimization problems, from setting up variables to verifying maxima and minima.
优化是微积分的一个基本应用，它使我们能够找到最佳结果，例如最大利润或最低成本。在IB数学中，优化问题常出现在试卷二中，要求学生建立现实世界场景的模型并应用微分技巧。本文提供了求解优化问题的全面指南，从设立变量到验证最大值和最小值。
1. What is Optimization? | 什么是优化？
Optimization is the process of maximizing or minimizing a quantity subject to given constraints. For example, we might wish to maximize the area of a rectangle with a fixed perimeter, or minimize the surface area of a cylinder of fixed volume. In calculus, we use the derivative to locate critical points where the rate of change is zero, and then determine whether these points correspond to maxima or minima.
优化是在给定约束条件下最大化或最小化某个量的过程。例如，我们可能希望最大化固定周长的矩形面积，或最小化固定体积的圆柱体的表面积。在微积分中，我们利用导数来找到变化率为零的临界点，然后确定这些点对应最大值还是最小值。
In IB Mathematics: Analysis and Approaches, optimization appears in the topic of calculus, both in SL and HL. Students must be comfortable with differentiation rules and algebraic manipulation. The key idea is to express the quantity to be optimized as a function of a single variable using the constraint, then differentiate.
在IB数学：分析与方法中，优化出现在微积分专题，SL和HL都涉及。学生必须熟练掌握求导法则和代数运算。核心思想是利用约束条件将需优化的量表示为单变量函数，然后求导。
2. General Strategy for Optimization | 优化的一般策略
To systematically tackle an optimization problem, follow these steps: 1) Identify the quantity Q to be optimized (objective function). 2) Draw a diagram if possible and label variables. 3) Write Q as a function in terms of one variable using given constraints. 4) Determine the domain of the variable (practical and mathematical). 5) Find the derivative dQ/dx and set it equal to zero to find critical points. 6) Verify which critical point(s) yield the desired optimum using the second derivative test or a sign chart. 7) State the answer in the context of the problem with correct units.
要系统地解决优化问题，请遵循以下步骤：1) 确定要优化的量 Q（目标函数）。2) 如果可能，画出示意图并标注变量。3) 利用给定的约束条件将 Q 表示为单个变量的函数。4) 确定变量的定义域（实际域和数学域）。5) 求导 dQ/dx 并令其为零，找到临界点。6) 使用二阶导数检验或符号表验证哪个临界点得到所需的最优值。7) 在问题上下文中给出答案，并带正确的单位。
The most common mistake is failing to express the objective function in terms of a single variable before differentiating. Always eliminate extra variables first by substituting the constraint equation. Pay attention to units and practical restrictions, such as lengths and volumes being positive.
最常见的错误是在求导前未能将目标函数表为单个变量的函数。务必先通过代入约束方程消除多余变量。注意单位和实际限制，如长度和体积必须为正。
3. Example: Maximizing Area of a Rectangular Pen | 示例：最大化矩形围栏面积
A farmer has 100 m of fencing to enclose a rectangular pen against an existing wall (only three sides need fencing). Find the dimensions that maximize the area.
一个农民有 100 米围栏，靠着一面已有的墙围一个矩形围栏（只需围三边）。求使面积最大的尺寸。
0 (since one length is the wall). Express y = 100 - 2x. Then A = x(100 - 2x) = 100x - 2x².
设宽为 x，长为 y，面积 A = x·y。约束：2x + y = 100（因一边是墙）。则 y = 100 - 2x，于是 A = x(100 - 2x) = 100x - 2x²。
A = 100x - 2x²
Domain: x > 0 and y > 0 ⇒ 100 - 2x > 0 ⇒ x < 50, so 0 < x < 50. Differentiate: dA/dx = 100 - 4x. Set to zero: 100 - 4x = 0 ⇒ x = 25.
定义域：x > 0 且 y > 0 ⇒ 100 - 2x > 0 ⇒ x < 50，所以 0 < x < 50。求导：dA/dx = 100 - 4x。令其为零：100 - 4x = 0 ⇒ x = 25。
-4 < 0, indicating a maximum. Then y = 100 - 2(25) = 50. Maximum area = 25 × 50 = 1250 m².
二阶导数：d²A/dx² = -4 < 0，表明为极大值。则 y = 100 - 2(25) = 50。最大面积 = 25 × 50 = 1250 m²。
4. Example: Minimizing Perimeter of a Rectangle with Fixed Area | 示例：固定面积下最小化矩形周长
t of fencing needed. Let length = L, width = W, area A = L·W = 200. Perimeter P = 2L + 2W.
假设我们想要一个面积为200平方米的矩形花园，并希望最小化所需的围栏长度。设长 = L，宽 = W，面积 A = L·W = 200。周长 P = 2L + 2W。
Express W = 200/L. Then P(L) = 2L + 400/L. Domain L > 0. Differentiate: P'(L) = 2 - 400/L². Set P'=0: 2 - 400/L² = 0 → L²
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