Локални екстремуми на функция

[luben_b]

Здравейте, задачата е следната:
Z= 10x^4 + 10y^4 - x^2 - y^2. Знам че се прави частна производна, но когато екстремума е за 2ра стерен, а не за 4та. Как да подходя?

[Добромир Глухаров]

$z=10x^4+10y^4-x^2-y^2$

$\begin{array}{|l}\frac{\partial z}{\partial x}=40x^3-2x=0\\\frac{\partial z}{\partial y}=40y^3-2y=0\end{array}$

$\begin{array}{|l}x(20x^2-1)=0\\y(20y^2-1)=0\end{array}$

$(x;y)\in\{(0;0),\left(0;\sqrt{\frac{1}{20}}\right),\left(0;-\sqrt{\frac{1}{20}}\right),\left(\sqrt{\frac{1}{20}};0\right),\left(\sqrt{\frac{1}{20}};\sqrt{\frac{1}{20}}\right),\left(\sqrt{\frac{1}{20}};-\sqrt{\frac{1}{20}}\right),\left(-\sqrt{\frac{1}{20}};0\right),\left(-\sqrt{\frac{1}{20}};\sqrt{\frac{1}{20}}\right),\left(-\sqrt{\frac{1}{20}};-\sqrt{\frac{1}{20}}\right)\}$

$\frac{\partial^2z}{\partial x^2}=120x^2-2$
$\frac{\partial^2z}{\partial y^2}=120y^2-2$
$\frac{\partial^2z}{\partial x\partial y}=0$

$\delta_{x,y}=\frac{\partial^2z}{\partial x^2}\cdot\frac{\partial^2z}{\partial y^2}-\left(\frac{\partial^2z}{\partial x\partial y}\right)^2=(120x^2-2).(120y^2-2)$

За да има екстремум, трябва $\delta_{x,y}>0$

$\Rightarrow x=\pm\sqrt{\frac{1}{20}}\wedge y=\pm\sqrt{\frac{1}{20}}\cup x=0\wedge y=0$

$(x;y)=\left(\pm\sqrt{\frac{1}{20}};\pm\sqrt{\frac{1}{20}}\right)\Rightarrow\frac{\partial^2 z}{\partial x^2}=120\cdot\frac{1}{20}-2>0\Rightarrow z_{min}=\frac{1}{40}+\frac{1}{40}-\frac{1}{20}-\frac{1}{20}=-\frac{1}{20}$

$(x;y)=(0;0)\Rightarrow\frac{\partial^2 z}{\partial x^2}=120.0-2<0\Rightarrow z_{max}=10.0+10.0-0-0=0$


https://www.wolframalpha.com/input/?i=z%3D10x%5E4%2B10y%5E4-x%5E2-y%5E2