Question

Найдите экстремум функции: z=2xy-2x^2-y^2+2y+4

Answer 1

$\displaystyle\begin{cases}z'_x=2y-4x=0\\+\\z'_y=2x-2y+2=0\end{cases}\\-2x+2=0\\x=1\\y=2\\z''_{xx}=-4\\z''_{yy}=-2\\z''_{xy}=2\\\mathcal4=AC-B^2=8-4=4;A<0\\z_{max}=4-2-4+4+4=6$