Answer
Local maximum at $f(3,\dfrac{3}{2})=\dfrac{17}{2}$
Work Step by Step
Since, we have $f_x(x,y)=2y-2x+3=0, f_y(x,y)=2x-4y=0$
After solving the above two equations, we get
$x=3,y=\dfrac{3}{2}$
Thus the critical point is: $(3,\dfrac{3}{2})$
As per second derivative test, we have
$D=f_{xx}f_{yy}-f^2_{xy}=4$ and $D=4 \gt 0$ and $f_{xx}=-2 \lt 0$
Thus, local maximum at $f(3,\dfrac{3}{2})=\dfrac{17}{2}$
