Answer
Local Minimum $f(\dfrac{1}{2}, 2 )=2 +\ln 2$
Work Step by Step
We have $f_x(x,y)=y+2-\dfrac{2}{x}; f_y(x,y)=x-y^{-1}=0$
Solving the above two equations leads to the critical point: $(\dfrac{1}{2}, 2 )$
Need, we take the help of second derivative test, for $(\dfrac{1}{2}, 2 )$.
$D=f_{xx}f_{yy}-f^2_{xy}=(8)(\dfrac{1}{4})-1=1 \gt 0$
So, the Local Minimum is
$f(\dfrac{1}{2}, 2 )=2-\ln \dfrac{1}{2}=2 +\ln 2$
