Question

Привет, мне нужна помощь 1 и 3 задание , желательно подробное решение

Answer 1

$1.;z(x,y)=frac{cos(x^2-y^2)}{3y}\ frac{partial z}{partial x}=-frac{2xsin(x^2-y^2)}{3y}\ frac{partial z}{partial y}=frac{-(-2y)sin(x^2-y^2)3y-3cos(x^2-y^2)}{9y^2}=frac{2y^2sin(x^2-y^2)3y-cos(x^2-y^2)}{3y^2}\ dz=left(-frac{2xsin(x^2-y^2)}{3y}right)dx+left(frac{2y^2sin(x^2-y^2)3y-cos(x^2-y^2)}{3y^2}right)dy$

$2.;z(x,y)=x^3-5x^2y^3+y^3+3y\ frac{partial z}{partial x}=3x^2-10xy^3\ frac{partial z}{partial y}=-15x^2y^2+3y+3\ frac{partial^2 z}{partial x^2}=6x-10y^3\ frac{partial^2 z}{partial y^2}=-30x^2y+3\ frac{partial^2 z}{partial xpartial y}=-30xy^2\ 3.;z(x,y)=x^2-4xy+y^2+12y-2\ begin{cases} frac{partial z}{partial x}=0\ frac{partial z}{partial y}=0 end{cases}Rightarrow begin{cases} 2x-4=0\ -4x+2y+12=0 end{cases}Rightarrow begin{cases} x=2\ y=-2 end{cases}$

$A=frac{partial^2 z}{partial x^2}=2\ B=frac{partial^2 z}{partial xpartial y}=-4\ C=frac{partial^2 z}{partial y^2}=2\ Delta=AcdotC-B^2=2cdot2-(-4)^2=4-16=-8<0$

Экстремума нет.