Answer
Local maximum of $f(\dfrac{1}{2},1)=-3-2 \ln 2$
Work Step by Step
Since, we have $f_x(x,y)=-4+\dfrac{2}{x}=0, f_y(x,y)=-1+\dfrac{1}{y}=0$
After solving the above two equations, we get
The critical point is: $(\dfrac{1}{2},1)$
As per the second derivative test, for $(\dfrac{1}{2},1)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=8$ and $D=8 \gt 0$ and $f_{xx}=-8 \lt 0$
Thus, we have: Local maximum of $f(\dfrac{1}{2},1)=-3-2 \ln 2$
Hence: Local maximum of $f(\dfrac{1}{2},1)=-3-2 \ln 2$
