## Thomas' Calculus 13th Edition

Local Minimum: $f(\dfrac{1}{2}, 2 )=2-\ln \dfrac{1}{2}=2 +\ln 2$
$$f_x(x,y)=y+2-\dfrac{2}{x} \\ f_y(x,y)=x-\dfrac{1}{y}=0$$ The critical points are: $(\dfrac{1}{2}, 2 )$ Apply the second derivative test for the critical point $(\dfrac{1}{2}, 2 )$. $D(\dfrac{1}{2}, 2 )=f_{xx}f_{yy}-f^2_{xy}=(8)(\dfrac{1}{4})-1=2-1=1 \gt 0$ Now, we have the Local Minimum: $f(\dfrac{1}{2}, 2 )=2-\ln \dfrac{1}{2}=2 +\ln 2$