柯西不等式最大值

Question

已知2x+3y=4
求8/x+27/y的最大值
用柯西不等式求解

Réflexion

Setting up the equation

I’m working on maximizing $8/x + 27/y$ under the constraint $2x + 3y = 4$ . Assuming $x > 0$ and $y > 0$ , this ensures the expression stays positive and well-defined.

Analyzing the function

As $x$ nears zero, the expression $8/x + 27/y$ becomes infinitely large. Since the constraints are $x > 0$ and $y > 0$ , a maximum may not exist.

Taking a closer look

I'm analyzing the situation using the Cauchy-Schwarz inequality to handle the constraint $2x + 3y = 4$ , ensuring $x$ and $y$ stay positive and the expression remains finite.

Examining wider constraints

OK, let me see. I’m considering if there might be extra restrictions on $x$ and $y$ , like them being negative, which could impact the maximum of $8/x + 27/y$ .

Examining finite constraints

No finite maximum exists for $8/x + 27/y$ . Considering $x$ and $y$ as real numbers without sign constraints, I’ll differentiate to find potential maximum values.

Taking a closer look

I’m differentiating $f(x) = 8/x + 81/(4-2x)$ with respect to $x$ , excluding $x = 0$ and $x = 2$ , to identify critical points where the derivative equals zero.

Applying the quadratic formula

I’m solving for $x$ by equating and simplifying the derivative $f'(x)$ to zero, using the quadratic formula to find $x = 8/13$ or $x = -1.6$ .

Confirming the solution

I’m verifying that $x = 8/13$ and $y = 12/13$ satisfy the constraint $2x + 3y = 4$ , and calculating the expression $8/x + 27/y$ to ensure the calculated value is correct.

Confirming the outcome

OK, let me confirm, $x = -1.6$ gives a negative result. The maximum value of $8/x + 27/y$ comes from $x = 8/13$ and $y = 12/13$ , totaling 42.25.

Noticing the contradiction

It's interesting to see that $8/x + 27/y$ is unbounded when $x$ or $y$ approach zero. The question might need reconsideration or domain constraints adjustments.

Revisiting constraints

I'm realizing the question might require using the Cauchy-Schwarz inequality to pinpoint the minimum of $8/x + 27/y$ under $2x + 3y = 4$ , with $x$ and $y$ both positive.

Analyzing the function

Interestingly enough, the expression $8/x + 27/y$ can grow indefinitely as $x$ or $y$ approach zero. The maximum of this function with both variables positive is unbounded.

Pinpointing the optimal value

The function $8/x + 27/y$ lacks an upper bound as $x$ or $y$ approach zero, suggesting that the critical point $8/13$ and $y = 12/13$ is possibly the minimum.

Breaking down Cauchy-Schwarz

I’m working through the Cauchy-Schwarz inequality, adjusting it to capture the constraint $2x + 3y = 4$ . This helps identify that the minimum of $8/x + 27/y$ is 42.25.

Evaluating constraints

I’m exploring the Cauchy–Schwarz inequality and confirming the minimum value of $8/x + 27/y$ as 42.25, with the function being unbounded above as $x$ or $y$ approach zero.

Reaching conclusions

I'm concluding that the function $8/x + 27/y$ is unbounded as $x$ or $y$ approach zero, suggesting the question might be misstated or focused on the minimum value.

Pulling together

OK, I’m working through the infinite unboundness of $8/x + 27/y$ as $x$ or $y$ approach zero, suggesting the question might aim to pinpoint the minimum value.

Confirming the unbounded nature

I’m concluding that $8/x + 27/y$ can grow without limit as $x$ or $y$ approach zero, indicating an unbounded function and possibly an error in the question.

Confirming the outcome

Interestingly enough, using the Cauchy-Schwarz inequality, I verified that $8/x + 27/y$ attains a minimum of 42.25. As $x$ or $y$ approach zero, the function is unbounded above.

Thought about 柯西不等式应用，持续 1m 5s

Réponse

先澄清一个要点：在通常的高中/大学竞赛或练习题背景下，若题目给出线性约束

2x + 3y \;=\; 4

并要求极大化

\dfrac{8}{x} \;+\;\dfrac{27}{y},

往往还会隐含「 $x>0$ 且 $y>0$ 」这样的条件(使得分母不为零、表达式可正则讨论)。但一旦 $x,y>0$ ，你会发现：

当 $x \to 0^+$ 时，由于 $2x + 3y = 4$ ，可得 $y = \tfrac{4}{3}$ ，这时 $\tfrac{8}{x} \to +\infty$ 。
当 $y \to 0^+$ 时，类似地 $x = 2$ ，这时 $\tfrac{27}{y} \to +\infty$ 。

因此

\frac{8}{x} + \frac{27}{y}

在可行域( $x>0,\,y>0$ )下其实是「没有上界」的，也就是说它可以无限大，不存在有限的“最大值”。如果题目文字的确是“求最大值”，那么严格来说答案应是：「此表达式在所给条件下无上界，最大值为 $+\infty$ 」。

常见的“柯西不等式”套路其实是在求下界

很多同学见到“用柯西不等式”会想到去套

\bigl(\tfrac{8}{x} + \tfrac{27}{y}\bigr)\,\bigl(2x + 3y\bigr) \; \ge \; \bigl(\sqrt{\tfrac{8}{x}\cdot 2x} \;+\; \sqrt{\tfrac{27}{y}\cdot 3y}\bigr)^2

从而推得

\bigl(\tfrac{8}{x} + \tfrac{27}{y}\bigr) \cdot 4 \;\;\ge\;\; \bigl(\sqrt{16} + \sqrt{81}\bigr)^2 \;=\; (4 + 9)^2 \;=\; 169,

即

\tfrac{8}{x} + \tfrac{27}{y} \;\;\ge\;\; \dfrac{169}{4}.

这个不等式给出的是下界（最小值），而不是上界（最大值）。进一步结合求导或配合“等号条件”可以验证，当且仅当

\sqrt{\dfrac{8}{x}}\;\big/\;\sqrt{\dfrac{27}{y}} \;=\; \sqrt{\dfrac{2x}{3y}}

时取等号，解得

x \;=\;\dfrac{8}{13}, \quad y \;=\;\dfrac{12}{13},

并且此时

\tfrac{8}{x} + \tfrac{27}{y} \;=\; \dfrac{169}{4} \;=\; 42.25.

也就是说，在 $x,y>0$ 并满足 $2x + 3y = 4$ 的所有可行对 $(x,y)$ 里， $\tfrac{8}{x} + \tfrac{27}{y}$ 的最小值是 $169/4$ 。但它可以通过令 $x$ 或 $y$ 趋近于 0 而变得无限大，故无最大值。

小结

若题意真的是“求最大值”，在典型的「 $x,y>0$ 且 $2x + 3y = 4$ 」约束下， $\tfrac{8}{x} + \tfrac{27}{y}$ 向 $+\infty$ 无上界，所以答案是「最大值不存在（或为 $+\infty$ ）」。
若其实想要用柯西不等式“求值”，常见的是找它的最小值。用上述柯西或“配方”方法，很容易得到最小值是

\min \Bigl(\tfrac{8}{x} + \tfrac{27}{y}\Bigr) \;=\; \dfrac{169}{4},

并在 $\bigl(\tfrac{8}{13},\,\tfrac{12}{13}\bigr)$ 处取到。

很多练习题其实是想让你证明

\tfrac{8}{x} + \tfrac{27}{y} \;\;\ge\;\; \dfrac{169}{4},

而不小心在措辞上写成“最大值”，需要分辨清楚题目的真实意图。若严格按字面意思回答，那么“最大值”确实是不存在的(无穷大)。如果你在课本或试卷里看到类似用柯西不等式推到 169/4，然后命题说“由此可见其最小值是 169/4”，这才是通常出题的初衷。

结论：

在正数范围内，本题所求 $\frac{8}{x} + \frac{27}{y}$ 没有有限的最大值，能够趋向 $+\infty$ 。
如果题目其实想问下界(最小值)，则答案是 $\displaystyle \frac{169}{4}$ 。

Question

Réflexion

Réponse

常见的“柯西不等式”套路其实是在求下界

小结

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