Solving Optimization Problems · 求解优化问题
From setup to the winning number
- You've reduced the problem to one variable. Now solve the optimization.
- The plan: differentiate the objective, find critical points, and confirm which is the max or min.
- Then answer the actual question — often the dimensions or the optimal value, with units.
- It's the extremum machinery of this unit, applied to a story.
从建模到那个制胜的数
- 你已把问题化到一个变量。现在求解这个最优化。
- 计划:对目标求导,找临界点,确认哪个是最大或最小。
- 然后回答真正的问题——常是尺寸或最优值,带单位。
- 这是本单元的极值机器,用在一个故事上。
The solving steps
- 1. Differentiate the single-variable objective and set it to zero to find critical points.
- 2. Confirm max vs. min with the first or second derivative test.
- 3. Check the endpoints of the domain too, if the domain is closed or bounded.
- 4. Answer the question asked — plug the optimal variable back in.
求解步骤
- 1. 对单变量目标求导并令其为零,找临界点。
- 2. 用一阶或二阶导数检验确认最大还是最小。
- 3. 若定义域是闭的或有界的,也检查定义域的端点。
- 4. 回答所问——把最优变量代回。
The area has a single peak · 面积有一个单一峰值
y = ax² + bx
The area parabola opens downward with a single peak — the derivative is zero at the vertex, a maximum. · 面积抛物线开口向下，有一个单一峰值——顶点处导数为零，是最大值。
For · 支持 $A(x)=40x-2x^2$, solve $A'(x)=40-4x=0$ for the critical $x$. · 对于$A(x)=40x-2x^2$，解$A'(x)=40-4x=0$求临界$x$。
$40-4x=0\Rightarrow x=10$.
Order the optimization solving steps. · 排列优化求解步骤。
Differentiate, find, confirm, answer. · 求导、查找、确认、回答。
Confirm, don't assume
- A critical point is only a candidate — you must show it's the max (or min) you want.
- Second Derivative Test: if $A''(x)<0$ at the critical point, it's a maximum.
- Or check the sign of $A'$ around it, or compare against domain endpoints.
- Skipping the confirmation is a common lost-mark; the AP expects a justification.
要确认,别假设
- 临界点只是候选——你必须证明它是你想要的最大(或最小)。
- 二阶导数检验: 若临界点处 $A''(x)<0$,它是最大值。
- 或检查 $A'$ 在其周围的符号,或与定义域端点比较。
- 跳过确认是常见的失分;AP 期望一个理由说明。
Since $A''(x)=-4<0$, the critical point is a... · 由于$A''(x)=-4<0$，临界点是一个...
Concave down ⇒ maximum. · 凹向下 ⇒ 最大值。
You should justify that a critical point is truly the maximum, not just assume it. · 你应该论证临界点确实是最大值，而不仅仅是假设它。
A derivative test or endpoint check is expected. · 预期进行导数测试或端点检查。
Answer the actual question
- If asked for the maximum area, report the area value; if asked for dimensions, report the lengths.
- Include units and check the answer is physically reasonable.
- Reread the prompt — "what dimensions?" and "what's the largest area?" want different final numbers.
- A number without the right label loses the point.
回答真正的问题
- 若问最大面积,报告面积值;若问尺寸,报告长度。
- 带上单位,并检查答案在物理上是否合理。
- 重读题目——"什么尺寸?"与"最大面积是多少?"要的是不同的最终数字。
- 没有正确标签的数字会失分。
With $x=10$ and $A(x)=40x-2x^2$, find the maximum area. · 给定$x=10$和$A(x)=40x-2x^2$，求最大面积。
$A(10)=400-200=200$.
If the question asks for the dimensions, you should report... · 如果问题要求尺寸，你应该报告...
Report what is asked — dimensions, not the area, here. · 报告所问内容——这里是尺寸，不是面积。
Two common slips: (1) reporting the critical $x$-value when the question wants the optimized quantity (or vice versa) — reread what's asked. (2) Skipping the justification that your critical point is truly the max/min. Always confirm with a derivative test or endpoint check.
两个常见失误:(1) 题目要优化后的量你却报告临界**$x$ 值**(或反之)——重读所问。(2) 跳过你的临界点确实是最大/最小的理由说明。永远用导数检验或端点比较来确认。
Maximize area $A(x)=40x-2x^2$ (from the fence setup), $0
- $A'(x)=40-4x=0\Rightarrow x=10$.
- $A''(x)=-4<0$ → concave down → a maximum ✓.
- Then $y=40-2(10)=20$, and $A=10\cdot20=200\ \text{m}^2$. Dimensions $10\times20$, max area $200\ \text{m}^2$.
最大化面积 $A(x)=40x-2x^2$(来自栅栏建模),$0
- $A'(x)=40-4x=0\Rightarrow x=10$。
- $A''(x)=-4<0$ → 下凹 → 一个最大值 ✓。
- 于是 $y=40-2(10)=20$,$A=10\cdot20=200\ \text{m}^2$。尺寸 $10\times20$,最大面积 $200\ \text{m}^2$。
To solve an optimization: differentiate the one-variable objective, set it to zero for critical points, confirm the max/min with a derivative test (and check domain endpoints), then answer the exact question with units. Don't report the wrong quantity, and always justify.
要求解最优化:对单变量目标求导,令其为零得临界点,用导数检验确认最大/最小(并检查定义域端点),然后带单位回答所问的确切问题。别报告错误的量,并永远给出理由。