Answer
Local maximum of $f(\dfrac{2}{3},\dfrac{4}{3})=0$
Work Step by Step
Since, we have $f_x(x,y)=2y-10x+4=0, f_y(x,y)=2x-4y+4=0$
After solving the above two equations, we get
$x=\dfrac{2}{3},y=\dfrac{4}{3}$
Thus the critical point is: $(\dfrac{2}{3},\dfrac{4}{3})$
As per second derivative test, we have
$D=f_{xx}f_{yy}-f^2_{xy}=36$ and $D=36 \gt 0$ and $f_{xx}(\dfrac{2}{3},\dfrac{4}{3})=-10 \lt 0$
Thus, local maximum of $f(\dfrac{2}{3},\dfrac{4}{3})=0$
