Finding minimum and maximum value of 3D function

Question

I have to find minimum and maximum of $$f(x,y)=\frac{x}{x^{2}+y^{2}+1}$$ on annulus $\left \{ (x,y): \epsilon ,: \mathbb{R}^{2}: :: 1\leq x^{2}+y^{2}\leq 4 \right \} $ From $\dfrac{\partial f}{\partial x}=\frac{-x^{2}+y^{2}+1}{(x^{2}+y^{2}+1)^{2}} =0$ and $\dfrac{\partial f}{\partial y}=\frac{-2xy}{(x^{2}+y^{2}+1)^{2}} =0$, I got two points $T_{1}=(1,0)$ and $T_{2}=(-1,0)$. When inserted in the original function I get two values of $\pm \dfrac{1}{2}$. How do I proceed from here?

Answer 1

The points $T_1$ and $T_2$ are just critical points of $f$. We can use the second derivative test to determine if either point is a minimum or a maximum you need to consider the $f_{xx}f_{yy}-f_{xy}^2$ evaluated at $T_1$ and $T_2$ to determine if either point is minimum or a maximum. If $f_{xx}(T_i)f_{yy}(T_i)-f_{xy}^2(T_i) > 0$ then $T$ is either a minimum or a maximum. If $f_{xx}(T_i)>0$ then we have a local minimum and if $f_{xx}(T_i) < 0$ we have a local maximum. You can read more here https://en.wikipedia.org/wiki/Hessian_matrix#Second-derivative_test