5. Znajdź i określ ekstrema funkcji

[Januszgolenia]

\(f(x,y)=x^2y+5ln(y^2)+y+6x\)

[eresh]

\(f(x,y)=x^2y+5ln(y^2)+y+6x\)

\(\frac{\partial f}{\partial x}=2xy+6\\
\frac{\partial f}{\partial y}=x^2+\frac{10y}{y^2}+1\\
\begin{cases}2xy+6=0\\x^2+\frac{10}{y}+1=0\end{cases}\\
\begin{cases}x=3\\y=-1\end{cases}\;\;\vee\;\;\begin{cases}x=\frac{1}{3}\\y=-9\end{cases}\\
\frac{\partial^2f}{\partial x^2}=2y\\
\frac{\partial ^2f}{\partial y^2}=-\frac{10}{y^2}\\
\frac{\partial^2f}{\partial x\partial y}=2x\\\)
\(\begin{bmatrix} \frac{\partial^2f}{\partial x^2}(3,-1)& \frac{\partial^2f}{\partial x\partial y}(3,-1)\\ \frac{\partial^2f}{\partial y\partial x}(3,-1)&\frac{\partial^2f}{\partial y^2}(3,-1)\end{bmatrix}=\begin{bmatrix} -2 & 6\\ 6&-10\end{bmatrix}<0\mbox{ brak ekstremum}\\
\begin{bmatrix} \frac{\partial^2f}{\partial x^2}(\frac{1}{3},-9)& \frac{\partial^2f}{\partial x\partial y}(\frac{1}{3},-9)\\ \frac{\partial^2f}{\partial y\partial x}(\frac{1}{3},-9)&\frac{\partial^2f}{\partial y^2}(\frac{1}{3},-9)\end{bmatrix}=\begin{bmatrix} -18 & \frac{2}{3}\\ \frac{2}{3}&-\frac{10}{81}\end{bmatrix}>0\;\;\wedge\;\; \frac{\partial^2f}{\partial x^2}(\frac{1}{3},-9)<0\mbox{ - maksimum}
\)