最优化问题简介

AP 微积分 AB· AP 微积分 AB CED — 微分的解析应用· 14 分钟阅读

1. 最优化问题核心概述★★☆☆☆⏱ 3 min

最优化问题是应用型微积分问题，要求你在给定一系列固定限制条件下，求某个实际量（如面积、成本、利润或体积）的最大或最小可能值。该知识点占AP微积分AB考试总分的15–18%，同时出现在选择题和自由作答题部分。

考试题目中，最优化的同义表述包括“求最大可能值”、“确定最小值”或“何种尺寸能最小化/优化”目标量。和抽象极值问题不同，最优化要求你首先根据文字描述构造需要优化的函数，这是AP考试中考查最多的技能。

📘 定义

最优化问题

需要在一组固定限制条件下求某个量的最大值或最小值的应用问题

例:

在围栏总长度固定的情况下，求能使面积最大的花园尺寸

2. 识别目标函数和约束函数★★★☆☆⏱ 3 min

任何最优化问题第一步，也是最容易出错的一步，就是将问题信息拆分为两个核心关系：

📘 定义

目标函数

问题要求你最大化或最小化其值的函数

例:

如果题目要求找使材料成本最小的尺寸，那么成本就是目标函数

📘 定义

约束函数

关联目标中多个变量的固定限制条件，可将目标简化为可求导的单变量函数

给所有未知量标注变量
写出需要优化的量的方程（即目标函数）
写出关联变量的固定约束方程
从约束方程解出一个变量，代入目标函数得到单变量函数

📐 例题

一个矩形花园，用总长120英尺的围栏围起来，花园一边邻河，沿河一侧不需要围栏。请识别目标函数和约束函数，再将目标函数化简为单变量函数，以最大化花园面积。

1
Let $x$ = length of each side perpendicular to the river, and $y$ = length of the side parallel to the river. Both are positive lengths, so $x > 0, y > 0$ .
2
We need to maximize area, so the objective function is:
3
$A = x \cdot y$
4
The total fencing available is 120 feet, with fencing only for two perpendicular sides and one parallel side, so the constraint is:
5
$2x + y = 120$
6
Solve the constraint for $y = 120 - 2 x$ , then substitute into the objective to get the single-variable objective:
7
$A(x) = x(120 - 2x) = 120x - 2x^2$

Exam tip:

所有几何最优化问题都要画带标注的示意图，避免建立模型出错

3. 求有效实际定义域★★★☆☆⏱ 3 min

Once you have a single-variable objective function, the next step is to find the interval of $x$ -values that make physical sense in the problem’s context. This domain is almost always a closed interval, so the Extreme Value Theorem applies: absolute extrema will occur either at critical points inside the interval or at the endpoints of the interval.

A common mistake is using the algebraic domain of the objective function instead of the contextual domain. For example, a quadratic objective function has an algebraic domain of all real numbers, but length cannot be negative or larger than the total amount of material available. To find the contextual domain:

Require all original variables to be non-negative (zero is allowed for endpoints)
Write inequalities for each variable, then solve for your single variable to get bounds
Confirm the final domain is a closed interval

📐 例题

Given the single-variable area objective function $A (x) = 120 x - 2 x^{2}$ from the garden example, find the valid contextual domain for $x$ .

1
Original variables: $x$ must be non-negative, so $x \geq 0$ . The parallel length $y = 120 - 2 x$ must also be non-negative.
2
Solve the inequality for $y$ :
3
$120 - 2x \geq 0 \implies x \leq 60$
4
Combine the bounds to get the valid closed domain:
5
$x \in [0, 60]$

Exam tip:

跳过定义域建立步骤会扣分，即使你的导数计算正确

4. Finding Optimal Values and Interpreting Results★★★★☆⏱ 5 min

The final step of the optimization process is testing to find the absolute maximum or minimum, then answering the question asked. For a closed interval domain, follow this process:

Compute the first derivative of the objective function
Find all critical points that lie inside the domain
Evaluate the objective function at every interior critical point and both endpoints
Select the largest value for a maximum, or the smallest for a minimum, then find any other requested values using the constraint

If the domain is open (e.g., $r > 0$ for a radius) and there is only one critical point, you can use the second derivative test: if $f^{''} (c) < 0$ for all $x$ in the domain, the critical point is an absolute maximum; if $f^{''} (c) > 0$ , it is an absolute minimum. This works for most open-domain AP optimization problems.

📐 例题

For the rectangular garden problem (maximize area with 120 feet of fencing, no fencing along the river), find the maximum area and the dimensions that produce it.

1
We start with the objective and domain from previous steps:
2
$A(x) = 120x - 2x^2, \quad x \in [0, 60]$
3
Compute the first derivative and find critical points:
4
$A'(x) = 120 - 4x \\ 120 - 4x = 0 \implies x = 30$
5
$x = 30$ is inside the domain. Evaluate $A (x)$ at all candidate points:
6
$A(0) = 0, \quad A(30) = 1800, \quad A(60) = 0$
7
The maximum value is 1800 square feet at $x = 30$ . Use the constraint to find $y$ :
8
$y = 120 - 2(30) = 60$
9
Final answer: Dimensions 30 ft (perpendicular) by 60 ft (parallel), maximum area 1800 square feet.

✓ 快速检测

Test your understanding of setting up objective functions:

A rectangular box with a square base is to be constructed. The total surface area of all 6 sides is 150 square inches. If $x$ is the side length of the square base, which of the following is the correct single-variable objective function for maximizing the volume of the box?
- $V (x) = x^{2} (150 - 2 x^{2})$
- $V (x) = \frac{150 x - 2 x ^{2}}{4 x}$
- $V (x) = \frac{150 x - 2 x ^{3}}{4}$
- $V (x) = 150 x - 6 x^{2}$
显示答案
2 —
Correct. Let $h$ = height. Surface area constraint $2 x^{2} + 4 x h = 150$ solves to $h = \frac{150 - 2 x ^{2}}{4 x}$ . Substitute into $V = x^{2} h$ to get the result.

📐 例题

An open-top rectangular box is made by cutting out squares of equal side length $x$ from each corner of a 12 inch by 18 inch sheet of cardboard, then folding up the sides. Find the objective function, domain, critical points, and maximum volume.

1
Height of the box equals $x$ , and base dimensions after cutting are $(12 - 2 x)$ and $(18 - 2 x)$ . The volume objective is:
2
$V(x) = x(12 - 2x)(18 - 2x)$
3
All dimensions must be non-negative, so domain is:
4
$x \in [0, 6]$
5
Expand, differentiate, and solve for critical points:
6
$V(x) = 4x^3 - 60x^2 + 216x \\ V'(x) = 12(x^2 - 10x + 18) \\ x = 5 \pm \sqrt{7}$
7
Only $x = 5 - 7 \approx 2.35$ is inside the domain. Second derivative test confirms it is a maximum:
8
$V''(5 - \sqrt{7}) = -24\sqrt{7} < 0$
9
Evaluate at candidates: $V (0) = 0$ , $V (6) = 0$ , $V (2.35) \approx 228$ . Final answer: Maximum volume is approximately 228 cubic inches at $x \approx 2.35$ inches.

📐 例题

A bakery's cost for a 6-inch tall round cake is $C (r) = 0.1 r^{2} + \frac{10}{r} + 2$ dollars, where $r$ is the radius in inches, $r > 0$ . What radius minimizes cost, and what is the minimum cost?

1
This is an open-domain problem with one critical point. Compute first derivative and set to zero:
2
$C'(r) = 0.2r - \frac{10}{r^2} \\ 0.2r = \frac{10}{r^2} \implies r^3 = 50 \implies r \approx 3.68$
3
Second derivative is always positive for $r > 0$ , so this is an absolute minimum:
4
$C''(r) = 0.2 + \frac{20}{r^3} > 0 \quad \forall r>0$
5
Evaluate cost: $C (3.68) \approx 6.07$ . Final answer: A radius of ~3.7 inches minimizes cost, for a minimum cost of ~$6.07 per cake.

5. 常见陷阱

错误做法:

Writing the constraint as $2 x + 2 y = 120$ instead of $2 x + y = 120$ for the river-side garden problem.

原因:

Students default to memorized full perimeter formulas instead of reading that one side needs no fencing.

正确做法:

Label every side that requires material on your diagram, then add up lengths explicitly instead of relying on memorized formulas.

错误做法:

Using the algebraic domain of the objective function instead of the contextual domain, e.g., leaving the domain as $(- \infty, \infty)$ for the garden problem.

原因:

Students forget that length cannot be negative, and context restricts variables to a physically meaningful interval.

正确做法:

After writing the single-variable objective, always write inequalities requiring all original variables to be non-negative, then solve to get the closed contextual domain.

错误做法:

Stopping after finding the critical point and not evaluating the objective function at endpoints.

原因:

Students assume the critical point must be the extremum and forget the Extreme Value Theorem requires checking endpoints.

正确做法:

Always list all interior critical points and both endpoints, evaluate the objective at all candidates, then select the maximum/minimum.

错误做法:

After finding the optimal $x$ , forgetting to compute other requested variables from the constraint, e.g., giving only $x = 30$ as the answer to a dimension question.

原因:

Students stop early after finding the critical point and do not confirm what the question asks for.

正确做法:

Reread the question's last sentence before writing your final answer to confirm you provided all requested values.

错误做法:

Justifying a local maximum instead of an absolute maximum.

原因:

Students confuse local and absolute extrema, and AP requires explicit justification of absolute extrema for optimization.

正确做法:

Either evaluate the objective at all candidates to show your value is the largest, or use the second derivative test to confirm concavity over the entire domain.

错误做法:

Solving the constraint for a squared variable, leading to square roots and unnecessary chain rule steps.

原因:

Students pick the first variable to solve for without checking which is easier.

正确做法:

Always solve the constraint for the variable that has a power of 1, to avoid roots and extra algebra errors.

6. 速查表

Category	Rule/Formula	Notes
Objective Function	$f (x)$ = quantity to maximize/minimize	Always single-variable after substituting constraint
Constraint Function	$g (x, y) = fixed constant$	Fixed limiting relationship between variables
Contextual Domain Rule	All original variables $\geq 0$	Endpoints included to form a closed interval
Closed Interval Optimization	Evaluate $f (x)$ at all interior critical points + endpoints	Largest = absolute maximum, smallest = absolute minimum
Open Domain (Single Critical Point)	$f^{''} (c) < 0 ⟹$ absolute max; $f^{''} (c) > 0 ⟹$ absolute min	Applies to $(0, \infty)$ domains with one critical point
Closed 6-sided Box (Square Base)	SA = $2 x^{2} + 4 x h$ , $V = x^{2} h$	$x$ = base side, $h$ = height
Cut-Out Open Box	$V (x) = x (L - 2 x) (W - 2 x)$	$x$ = cut-out side, $L, W$ = original sheet dimensions

真题中的出现

AI 根据考纲规律估算的考点位置,请对照官方真题核实准确性。仅作复习重点参考。

2023 · AB
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2022 · AB MCQ
识别正确的目标函数

下一步

Mastering the core 4-step optimization process is the foundation for all applied extremum problems in the AP Calculus AB curriculum. This topic ties together all prior Unit 5 concepts: derivatives, critical points, and extrema into a single applied framework that is heavily tested on both multiple-choice and free-response sections of the AP exam. Next, you will move on to more complex optimization problems, and connect these skills to related rates, where you optimize a changing quantity over time. You will also apply this same process to problems with non-linear constraints that require implicit differentiation. Without mastering this core process, more complex applied problems will be much harder to solve correctly.