дана функция z=x^2+3y^2-8x-2y-1 найти ее точки экстремума

$\\ z=x^2+3y^2-8x-2y-1\\ z'_x=2x-8\\ z'_y=6y-2\\\\$

$\\2x-8=0\\ 2x=8\\ x=4\\\\ 6y-2=0\\ 6y=2\\ y=\frac{1}{3}\\\\ (4,\frac{1}{3})$

$\\z''_{xx}=2\\ z''_{yy}=6\\ z''_{xy}=0\\ \Delta=2\cdot6-0^2=\\ \Delta=12\\$

0\wedge \Delta>0 \Rightarrow z(4,\frac{1}{3})=z_{min}" alt="\\z(4,\frac{1}{3})=4^2+3\cdot(\frac{1}{3})^2-8\cdot4-2\cdot(\frac{1}{3})-1\\ z(4,\frac{1}{3})=16+\frac{1}{3}-32-\frac{2}{3}-1\\ z(4,\frac{1}{3})=-17-\frac{1}{3}\\ z(4,\frac{1}{3})=-\frac{51}{3}-\frac{1}{3}\\ z(4,\frac{1}{3})=-\frac{52}{3}\\\\ z''_{xx}>0\wedge \Delta>0 \Rightarrow z(4,\frac{1}{3})=z_{min}" align="absmiddle" class="latex-formula">