Answer
Local maximum at $f(\dfrac{16}{7},0)=\dfrac{-16}{7}$
Work Step by Step
Since, we have $f_x(x,y)=\dfrac{112x-8x}{\sqrt{56x^2-8y^2-16x-31}}=0, f_y(x,y)=\dfrac{-8y}{\sqrt{56x^2-8y^2-16x-31}}=0$
After solving the above two equations, we get
$x=\dfrac{16}{7},y=0$
Thus the critical point is: $(\dfrac{16}{7},0)$
As per second derivative test, we have
$D=f_{xx}f_{yy}-f^2_{xy}=\dfrac{64}{225}$ and $D=\dfrac{64}{225} \gt 0$ and $f_{xx}=-\dfrac{-8}{15} \lt 0$
Thus, local maximum at $f(\dfrac{16}{7},0)=\dfrac{-16}{7}$
