Chapter 13 - Section 13.7 - Extreme Values and Saddle Points - Exercises - Page 737: 21

Local maximum of $f(0,0)=-1$

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