wyznacz ekstrema lokalne
: 19 sty 2022, 11:20
\(f(x,y)= - \ln (x^2+2y^2+e)\)
\(\frac{\partial f}{\partial x}=-\frac{2x}{x^2+2y^2+e}\\
\frac{\partial f}{\partial y}=-\frac{4y}{x^2+2y^2+e}\)
\(\begin{cases}-\frac{2x}{x^2+2y^2+e}=0\\-\frac{4y}{x^2+2y^2+e}=0\end{cases}\\
\begin{cases}x=0\\y=0\end{cases}\)
\(\frac{\partial^2f}{\partial x^2}=\frac{-2(x^2+2y^2+e)+4x^2}{(x^2+2y^2+e)^2}\\
\frac{\partial^2f}{\partial x\partial y}=\frac{8xy}{(x^2+2y^2+e)^2}\\
\frac{\partial^2f}{\partial y^2}=\frac{-4(x^2+2y^2+e)+16y^2}{(x^2+y^2+e)^2}\\\)
\(\frac{\partial^2f}{\partial x^2}(0,0)=-\frac{-2e}{e^2}=-\frac{2}{e}\\
\frac{\partial^2f}{\partial x\partial y}(0,0)=0\\
\frac{\partial^2f}{\partial y^2}(0,0)=-\frac{4}{e}\)
\(\begin{vmatrix}-\frac{2}{e}&0\\0&-\frac{4}{e} \end{vmatrix}=\frac{8}{e^2}>0\;\; \wedge\;\;\frac{\partial^2f}{\partial x^2}(0,0)<0 \)
maksimum w \((0,0)\)