题目
设f(x,y)与φ (x,y)均为可微函数,且 (varphi )_(y)'(x,y)neq 0. 已知-|||-(x0,y0)是f(x,y)在约束条件 varphi (x,y)=0 下的一个极值点,下列选项正确的是-|||-() .-|||-(A)若 _(x)'((x)_(0),(y)_(0))=0, 则 _(y)'((x)_(0),(y)_(0))=0.-|||-(B)若 _(x)'((x)_(0),(y)_(0))=0, 则 _(y)'((x)_(0),(y)_(0))neq 0.-|||-(C)若 _(x)'((x)_(0),(y)_(0))neq 0, 则 _(y)'((x)_(0),(y)_(0))=0.-|||-(D)若 _(x)'((x)_(0),(y)_(0))neq 0, 则 _(y)'((x)_(0),(y)_(0))neq 0.
题目解答
答案
解析
步骤 1：应用拉格朗日乘数法
根据拉格朗日乘数法，对于函数f(x,y)在约束条件φ(x,y)=0下的极值点(x0,y0)，存在一个常数λ，使得
\[
\left \{ \begin{matrix}
{f}_{x}'({x}_{0},{y}_{0})+{\lambda }{\varphi }_{x}'({x}_{0},{y}_{0})=0\\
{f}_{y}'({x}_{0},{y}_{0})+{\lambda }{\varphi }_{y}'({x}_{0},{y}_{0})=0
\end{matrix} \right.
\]
步骤 2：消去λ
从上述方程组中消去λ，得到
\[
{f}_{x}'({x}_{0},{y}_{0}){\varphi }_{y}'({x}_{0},{y}_{0})={f}_{y}'({x}_{0},{y}_{0}){\varphi }_{x}'({x}_{0},{y}_{0})
\]
步骤 3：分析条件
由于${\varphi }_{y}'(x,y)\neq 0$，可以将上式改写为
\[
{f}_{x}'({x}_{0},{y}_{0})=\dfrac {{f}_{y}'({x}_{0},{y}_{0}){\varphi }_{x}'({x}_{0},{y}_{0})}{{\varphi }_{y}'({x}_{0},{y}_{0})}
\]
步骤 4：判断选项
当${f}_{x}'({x}_{0},{y}_{0})\neq 0$时，根据上式，${f}_{y}'({x}_{0},{y}_{0})$不能为0，否则${f}_{x}'({x}_{0},{y}_{0})$将为0，与假设矛盾。因此，选项(D)正确。
根据拉格朗日乘数法，对于函数f(x,y)在约束条件φ(x,y)=0下的极值点(x0,y0)，存在一个常数λ，使得
\[
\left \{ \begin{matrix}
{f}_{x}'({x}_{0},{y}_{0})+{\lambda }{\varphi }_{x}'({x}_{0},{y}_{0})=0\\
{f}_{y}'({x}_{0},{y}_{0})+{\lambda }{\varphi }_{y}'({x}_{0},{y}_{0})=0
\end{matrix} \right.
\]
步骤 2：消去λ
从上述方程组中消去λ，得到
\[
{f}_{x}'({x}_{0},{y}_{0}){\varphi }_{y}'({x}_{0},{y}_{0})={f}_{y}'({x}_{0},{y}_{0}){\varphi }_{x}'({x}_{0},{y}_{0})
\]
步骤 3：分析条件
由于${\varphi }_{y}'(x,y)\neq 0$，可以将上式改写为
\[
{f}_{x}'({x}_{0},{y}_{0})=\dfrac {{f}_{y}'({x}_{0},{y}_{0}){\varphi }_{x}'({x}_{0},{y}_{0})}{{\varphi }_{y}'({x}_{0},{y}_{0})}
\]
步骤 4：判断选项
当${f}_{x}'({x}_{0},{y}_{0})\neq 0$时，根据上式，${f}_{y}'({x}_{0},{y}_{0})$不能为0，否则${f}_{x}'({x}_{0},{y}_{0})$将为0，与假设矛盾。因此，选项(D)正确。