Question

Решить с объяснением исследовать на экстремум: z=4x+5y-x^2-xy-y^2+4

Answer 1

$z=4x+5y-x^2-xy-y^2+4\\\\z'_{x}=4-2x-y=0\; ,\quad \; 2x+y=4\\\\z'_{y}=5-x-2y=0\; ,\quad \; x+2y=5\\\\ \left \{ {{2x+y=4} \atop {x+2y=5}} \right. \; \; \left \{ {{2x+y=4} \atop {-2x-4y=-10}} \right. \; \oplus \; \; \left \{ {{2x+y=4} \atop {-3y=-6}} \right. \; \; \; \left \{ {{x=1} \atop {y=2}} \right. \\\\M_0(1,2)\; \; stacionarnaya\; tochka\\\\z''_{xx}=-2\; ,\; \; z''_{xy}=-1\; ,\; \; z''_{yy}=-2\\\\A=z''_{xx}(M_0)=-2\; ,\; \; B=z''_{xy}(M_0)=-1\; ,\; \; C=z''_{yy}(M_0)=-2$

$\Delta =AC-B^2=-2\cdot (-2)-(-1)^2=4-1=30\; \; \to$

$\Delta \ \textgreater \ 0\; \; \Rightarrow \; \; \; est'\; \; extremym\\\\A=-2\ \textless \ 0\; \; \; \Rightarrow \; \; \; max\\\\z_{max}=z(1,2)=4\cdot 1+5\cdot 2-1^2-1\cdot 2-2^2+4=11$