91Ó°ÊÓ

A function,$f(x,y)=\frac{1}{x^2+y^2−1}$

$$\begin{gathered} \mathrm{The}\mathrm{first}-\mathrm{order}\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{are}: \\ {f}_x(x,y)=\frac{∂\mathrm{f}}{∂\mathrm{x}}=-\frac{2x}{{\left(x^2+y^2-1\right)}^2}\mathrm{and}{f}_y(x,y)=\frac{∂\mathrm{f}}{∂\mathrm{y}}=-\frac{2y}{{\left(x^2+y^2-1\right)}^2} \\ \mathrm{Now},\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}:-\frac{2x}{{\left(x^2+y^2-1\right)}^2}=0 \\ \mathrm{and}-\frac{2x}{{\left(x^2+y^2-1\right)}^2}=0,\mathrm{we}\mathrm{get}, \\ x=0\mathrm{and}y=0 \\ \mathrm{We}\mathrm{find}\mathrm{only}\mathrm{one}\mathrm{stationary}\mathrm{point}\mathrm{of}f,\mathrm{namely}:(0,0) \\ \mathrm{The}\mathrm{second}-\mathrm{order}\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{are}: \\ {f}_{xx}(x,y)=\frac{{∂}^2\mathrm{f}}{∂{\mathrm{x}}^2}=-\frac{2{\left(x^2+y^2-1\right)}^2-4x^2}{{\left({\left(x^2+y^2-1\right)}^2\right)}^2},f_{yy}(x,y)=\frac{{∂}^2\mathrm{f}}{∂{\mathrm{y}}^2}=-\frac{2{\left(x^2+y^2-1\right)}^2-4y^2}{{\left({\left(x^2+y^2-1\right)}^2\right)}^2} \\ \mathrm{and}{f}_{xy}(x,y)=\frac{{∂}^2\mathrm{f}}{∂\mathrm{x}∂\mathrm{y}}=\frac{8xy}{{\left(x^2+y^2-1\right)}^3} \\ {f}_{xx}(0,0)=-2,f_{yy}(0,0)=-2\mathrm{and}{f}_{xy}(0,0)=0 \\ \mathrm{The}\mathrm{determinant}\mathrm{of}\mathrm{the}\mathrm{Hessian}\mathrm{is}: \\ det\left(H_f\left(x,y\right)\right)=\left(\frac{{∂}^2\mathrm{f}}{∂{\mathrm{x}}^2}\right)\left(\frac{{∂}^2\mathrm{f}}{∂{\mathrm{y}}^2}\right)-{\left(\frac{{∂}^2\mathrm{f}}{∂\mathrm{x}∂\mathrm{y}}\right)}^2 \\ det\left(H_f\left(0,0\right)\right)=\left(-2\right)×\left(-2\right)-0^2=4 \end{gathered}$$

$$\begin{gathered} \mathrm{If}f\mathrm{has}\mathrm{a}\mathrm{stationary}\mathrm{point}\mathrm{at}({\mathrm{x}}_0,{\mathrm{y}}_0),\mathrm{then} \\ \left(\mathrm{a}\right)f\mathrm{has}\mathrm{a}\mathrm{relative}\mathrm{maximum}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_f\right(x_0,y_0\left)\right)>0\mathrm{with} \\ {f}_{xx}(x_0,y_0)<0or{f}_{yy}(x_0,y_0)<0. \\ \left(\mathrm{b}\right)f\mathrm{has}\mathrm{a}\mathrm{relative}\mathrm{minimum}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_f\right(x_0,y_0\left)\right)>0\mathrm{with} \\ {f}_{xx}(x_0,y_0)>0\mathrm{or}{f}_{yy}(x_0,y_0)>0. \\ \left(\mathrm{c}\right)f\mathrm{has}\mathrm{a}\mathrm{saddle}\mathrm{point}\mathrm{at}(x_0,y_0)\mathrm{if}det\left(H_f\right(x_0,y_0\left)\right)<0. \\ \left(\mathrm{d}\right)\mathrm{If}det\left(H_f\right(x_0,y_0\left)\right)=0,\mathrm{no}\mathrm{conclusion}\mathrm{may}\mathrm{be}\mathrm{drawn}\mathrm{about}\mathrm{the}\mathrm{behavior}\mathrm{of} \\ f\mathrm{at}(x_0,y_0). \\ \mathrm{In}\mathrm{the}\mathrm{given}\mathrm{function},det\left(H_f\left(0,0\right)\right)=4>0\mathrm{with}{f}_{xx}\left(0,0\right)=-2<0\mathrm{and}{f}_{yy}\left(0,0\right)=-2<0. \\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{maximum}\mathrm{at}\left(0,0\right)\mathrm{with}\mathrm{maximum}\mathrm{value}, \\ f\left(0,0\right)=\frac{1}{0^2+0^2-1}=-1 \end{gathered}$$

$$\begin{gathered} \mathrm{When}x=0,\underset{y→∞}{\lim }f(0,y)=0\mathrm{and}\underset{y→-∞}{\lim }f(0,y)=0 \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{local}\mathrm{maximum}\mathrm{at}\left(0,0\right). \\ \mathrm{There}\mathrm{are}\mathrm{no}\mathrm{absolute}\mathrm{maximum}\mathrm{and}\mathrm{absolute}\mathrm{minimnum}\mathrm{points}. \end{gathered}$$