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偏导数在求极值中的应用

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Versed_sine
Versed_sine
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前置知识​

“偏导数”一课通！1h零基础上手！|高数下​
信息

求导法则：

F(x)=f(x)±g(x),F′(x)=f′(x)±g′(x)F(x)=f(x)⋅g(x),F′(x)=f(x)⋅g′(x)+f′(x)⋅g(x)F(x)=f(x)g(x),F′(x)=f′(x)⋅g(x)+f(x)⋅g′(x)g2(x)F(x)=f[g(x)],F′(x)=f′(g(x))⋅g′(x)\begin{align*} & F(x) = f(x) \pm g(x), & F'(x) = f'(x) \pm g'(x) \\ & F(x) = f(x) \cdot g(x), & F'(x) = f(x) \cdot g'(x) + f'(x) \cdot g(x) \\ & F(x) = \frac{f(x)}{g(x)}, & F'(x) = \frac{f'(x) \cdot g(x) + f(x) \cdot g'(x)}{g^2(x)} \\ & F(x) = f[g(x)], & F'(x) = f'(g(x))\cdot g'(x) \end{align*}

例题​

不等式核心思想方法，分离与配凑？​

题面​

a,b>0a, b > 0，求 ab+ba2+b2+1\dfrac{ab+b}{a^2+b^2+1} 的最大值。

偏微分解法​

记 f(a,b)=ab+ba2+b2+1f(a, b) = \dfrac{ab+b}{a^2+b^2+1}

∂f∂a=b(a2+b2+1)2⋅[(a2+b2+1)−2a(a+1)]=b(a2+b2+1)2⋅(b2−a2−2a+1)\begin{aligned} \dfrac{\partial f}{\partial a} &= \dfrac{b}{(a^2+b^2+1)^2} \cdot [(a^2+b^2+1)-2a(a+1)] \\ &=\dfrac{b}{(a^2+b^2+1)^2} \cdot (b^2-a^2-2a+1) \end{aligned}

∂f∂b=a+1(a2+b2+1)2⋅[(a2+b2+1)−2b⋅b]=a+1(a2+b2+1)2⋅(a2−b2+1).\begin{aligned} \frac{\partial f}{\partial b} &= \frac{a+1}{(a^2+b^2+1)^2}\cdot[(a^2+b^2+1)-2b\cdot b] \\ &=\frac{a+1}{(a^2+b^2+1)^2}\cdot(a^2-b^2+1). \\ \end{aligned}

取极值时 ∂f∂a=∂f∂b=0\dfrac{\partial f}{\partial a} = \dfrac{\partial f}{\partial b} = 0，即： {b2−a2−2a+1=01◯a2−b2+1=02◯\begin{cases} b^2-a^2-2a+1=0 & \textcircled{\scriptsize 1} \\ a^2-b^2+1=0 & \textcircled{\scriptsize 2} \end{cases}

1◯+2◯:2−2a=0\textcircled{\scriptsize 1}+\textcircled{\scriptsize 2}: 2-2a=0，即 a=1a=1

易得 b=2b=\sqrt{2}

∴f(a,b)max⁡=f(1,2)=224=22\therefore f(a, b)_{\max} = f(1, \sqrt{2}) = \dfrac{2\sqrt{2}}{4} = \dfrac{\sqrt{2}}{2}