Chapter 13 - Partial Derivatives - Chapter 13 Review Exercises - Page 998: 34

$$\eqalign{ & {\text{relative maximum at }}\left( {0,0} \right) \cr & {\text{saddle point at }}\left( {6,3} \right) \cr & {\text{saddle point at }}\left( { - 6,3} \right) \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = {x^2}y - 6{y^2} - 3{x^2} \cr & \,\,\,\,\,{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2}y - 6{y^2} - 3{x^2}} \right] \cr & \,\,\,\,\,{f_x}\left( {x,y} \right) = 2xy - 6x \cr & \,\,\,\,\,{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2xy - 6x} \right] \cr & \,\,\,\,\,{f_{xx}}\left( {x,y} \right) = 2y - 6 \cr & and \cr & \,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2}y - 6{y^2} - 3{x^2}} \right] \cr & \,\,\,\,\,{f_y}\left( {x,y} \right) = {x^2} - 12y \cr & \,\,\,\,\,{f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - 12y} \right] \cr & \,\,\,\,\,{f_{yy}}\left( {x,y} \right) = - 12 \cr & {\text{Calculate the mixed partial derivative }}{f_{xy}}\left( {x,y} \right) \cr & \,\,\,\,\,{f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2xy - 6x} \right] \cr & \,\,\,\,\,{f_{xy}}\left( {x,y} \right) = 2x \cr & \cr & {\text{Setting the first partial derivatives to 0}} \cr & {f_x}\left( {x,y} \right) = 2xy - 6x = 0{\text{ and }}{f_y}\left( {x,y} \right) = {x^2} - 12y = 0 \cr & \,\,\,\,\,\,\,2x\left( {y - 3} \right) = 0 \cr & \,\,\,\,\,\,x = 0,\,\,\,\,\,\,y = 3 \cr & {\text{We have the equation }}{x^2} - 12y = 0.{\text{ for }}x = 0,\,\,\, \cr & \,\,\,\,\,\,\,\,{\left( 0 \right)^2} - 12y = 0\,\,\, \to \,\,\,\,y = 0,\,\,\,\,\,{P_1}\left( {0,0} \right) \cr & {\text{We have the equation }}{x^2} - 12y = 0.{\text{ for }}y = 3,\,\,\, \cr & \,\,\,\,\,\,\,{x^2} - 12\left( 3 \right) = 0 \cr & \,\,\,\,\,\,\,{x^2} = 36 \cr & \,\,\,\,\,\,\,x = \pm 6\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,{P_2}\left( {6,3} \right){\text{ and }}{P_3}\left( { - 6,3} \right) \cr & {\text{The critical points are: }}{P_1}\left( {0,0} \right),\,\,\,\,\,{P_2}\left( {6,3} \right),\,\,\,\,\,\,{P_3}\left( { - 6,3} \right) \cr & {\text{Use The Second Partials Test at the critical points}} \cr & D = {f_{xx}}\left( {{x_0},{y_0}} \right){f_{yy}}\left( {{x_0},{y_0}} \right) - f_{xy}^2\left( {{x_0},{y_0}} \right) \cr & D = \left( {2y - 6} \right)\left( { - 12} \right) - {\left( {2x} \right)^2} \cr & \cr & {\text{For }}{P_1}\left( {0,0} \right) \cr & D = \left( {2\left( 0 \right) - 6} \right)\left( { - 12} \right) - {\left( {2\left( 0 \right)} \right)^2} = 72,\,\,\,\,{f_{xx}}\left( {0,0} \right) = - 6 \cr & D > 0{\text{ and }}{f_{xx}} < 0,{\text{ There is a relative maximum at }}\left( {0,0} \right) \cr & \cr & {\text{For }}{P_2}\left( {6,3} \right) \cr & D = \left( {2\left( 3 \right) - 6} \right)\left( { - 12} \right) - {\left( {2\left( 6 \right)} \right)^2} = - 144 \cr & D < 0{\text{ }},{\text{ There is a saddle point at }}\left( {6,3} \right) \cr & \cr & {\text{For }}{P_3}\left( { - 6,3} \right) \cr & D = \left( {2\left( 3 \right) - 6} \right)\left( { - 12} \right) - {\left( {2\left( 6 \right)} \right)^2} = - 144 \cr & D < 0{\text{ }},{\text{ There is a saddle point at }}\left( { - 6,3} \right) \cr} $$