Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 8

$${\text{Saddle point at }}\left( {\frac{1}{2}, - \frac{1}{2},0} \right)$$

$$\eqalign{ & g\left( {x,y} \right) = {x^2} - {y^2} - x - y \cr & {\text{Calculate the first partial derivatives of }}g\left( {x,y} \right) \cr & {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} - {y^2} - x - y} \right] \cr & {g_x}\left( {x,y} \right) = 2x - 1 \cr & and \cr & {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - {y^2} - x - y} \right] \cr & {g_y}\left( {x,y} \right) = - 2y - 1 \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {g_x}\left( {x,y} \right) = 0 \cr & 2x - 1 = 0 \cr & x = \frac{1}{2} \cr & {g_y}\left( {x,y} \right) = 0 \cr & - 2y - 1 = 0 \cr & y = - \frac{1}{2} \cr & {\text{The critical point is }}\left( {\frac{1}{2}, - \frac{1}{2}} \right) \cr & {\text{Find the second partial derivatives of }}g\left( {x,y} \right){\text{ and }}{g_{xy}}\left( {x,y} \right) \cr & {g_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x - 1} \right] = 2 \cr & {g_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2y - 1} \right] = - 2 \cr & {g_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x - 1} \right] = 0 \cr & {\text{By the second partials test}} \cr & d\left( {x,y} \right) = {g_{xx}}\left( {a,b} \right){g_{yy}}\left( {a,b} \right) - {\left[ {{g_{xy}}\left( {x,y} \right)} \right]^2} \cr & {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {\frac{1}{2}, - \frac{1}{2}} \right) \cr & d\left( {\frac{1}{2}, - \frac{1}{2}} \right) = {g_{xx}}\left( {\frac{1}{2}, - \frac{1}{2}} \right){g_{yy}}\left( {\frac{1}{2}, - \frac{1}{2}} \right) - {\left[ {{g_{xy}}\left( {\frac{1}{2}, - \frac{1}{2}} \right)} \right]^2} \cr & d\left( {\frac{1}{2}, - \frac{1}{2}} \right) = \left( 2 \right)\left( { - 2} \right) - {\left[ 0 \right]^2} \cr & d\left( {\frac{1}{2}, - \frac{1}{2}} \right) = - 4 \cr & d < 0,{\text{ then}} \cr & g\left( {x,y} \right){\text{ has a saddle point at }}\left( {\frac{1}{2}, - \frac{1}{2},g\left( {\frac{1}{2}, - \frac{1}{2}} \right)} \right) \cr & g\left( {x,y} \right) = {x^2} - {y^2} - x - y \cr & g\left( {\frac{1}{2}, - \frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^2} - {\left( { - \frac{1}{2}} \right)^2} - \left( {\frac{1}{2}} \right) - \left( {\frac{{ - 1}}{2}} \right) = 0 \cr & {\text{Saddle point at }}\left( {\frac{1}{2}, - \frac{1}{2},0} \right) \cr} $$