Question

Z=x^2y^2-xy+y+x+2 экстремум функций

Answer 1

Найдём критические точки

$\begin{cases}\cfrac{\partial z}{\partial x}=2xy^2-y+1\\ \cfrac{\partial z}{\partial y}=2x^2y-x+1\end{cases}\Rightarrow \begin{cases}2xy^2-y+1=0\\ 2x^2y-x+1=0\end{cases}\Leftrightarrow \begin{cases}x=\cfrac{y-1}{2y^2}\\2x^2y-x+1=0\end{cases}\\1-\cfrac{y-1}{2y^2}+\cfrac{(y-1)^2}{2y^3}=0\Leftrightarrow 2y^3-y+1=0\Leftrightarrow (y+1)\underbrace{\left ( 2y^2-2y+1 \right )}_{y\in \mathbb{C}}=0\\y=-1\Rightarrow x=-1\Rightarrow M(-1,-1)$

Экстремумы

$\begin{cases}\cfrac{\partial ^2z}{\partial x^2}=2y^2\\ \cfrac{\partial ^2z}{\partial y^2}=2x^2\end{cases}\Rightarrow \cfrac{\partial ^2z}{\partial x \partial y}=4xy-1\\\cfrac{\partial ^2z}{\partial x^2}\Bigg|_{M}=2, \ \ \cfrac{\partial ^2z}{\partial y^2}\Bigg|_{M}=2, \ \ \cfrac{\partial ^2z}{\partial x \partial y}\Bigg|_{M}=3\\\cfrac{\partial ^2z}{\partial x^2}\cfrac{\partial ^2z}{\partial y^2}-\left ( \cfrac{\partial ^2z}{\partial x \partial y} \right )^2=-5 < 0$

Глобального экстремума нет!