Chapter 13 - Partial Derivatives - Chapter 13 Review Exercises - Page 998: 35

$$\eqalign{ & {\text{saddle point at }}\left( {0,0} \right) \cr & {\text{relative minimum at }}\left( {3,9} \right) \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = {x^3} - 3xy + \frac{1}{2}{y^2} \cr & \,\,\,\,\,{f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^3} - 3xy + \frac{1}{2}{y^2}} \right] \cr & \,\,\,\,\,{f_x}\left( {x,y} \right) = 3{x^2} - 3y \cr & \,\,\,\,\,{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^2} - 3y} \right] \cr & \,\,\,\,\,{f_{xx}}\left( {x,y} \right) = 6x \cr & and \cr & \,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^3} - 3xy + \frac{1}{2}{y^2}} \right] \cr & \,\,\,\,\,{f_y}\left( {x,y} \right) = - 3x + y \cr & \,\,\,\,\,{f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 3x + y} \right] \cr & \,\,\,\,\,{f_{yy}}\left( {x,y} \right) = 1 \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[ {3{x^2} - 3y} \right] \cr & \,\,\,\,\,{f_{xy}}\left( {x,y} \right) = - 3 \cr & \cr & {\text{Setting the first partial derivatives to 0}} \cr & {f_x}\left( {x,y} \right) = 3{x^2} - 3y = 0{\text{ and }}{f_y}\left( {x,y} \right) = - 3x + y = 0 \cr & \,\,\,\, - 3x + y = 0 \cr & \,\,\,\,\,\,\,y = 3x \cr & {\text{Substitute }}y = 3x{\text{ into }}3{x^2} - 3y = 0 \cr & \,\,\,\,\,\,\,3{x^2} - 3\left( {3x} \right) = 0 \cr & \,\,\,\,\,\,\,3{x^2} - 9x = 0 \cr & \,\,\,\,\,\,\,3x\left( {x - 3} \right) = 0 \cr & \,\,\,\,\,\,\,x = 0,\,\,\,\,and{\text{ }}x = 3 \cr & {\text{We have the equation }}y = 3x.{\text{ for }}x = 0 \to y = 0\,\,\, \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ for }}x = 3 \to y = 9\,\,\, \cr & {\text{The critical points are: }}{P_1}\left( {0,0} \right),\,\,\,\,\,and\,\,\,\,\,\,{P_2}\left( {3,9} \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( {6x} \right)\left( 1 \right) - {\left( { - 3} \right)^2} \cr & \cr & {\text{For }}{P_1}\left( {0,0} \right) \cr & D = \left( {6\left( 0 \right)} \right)\left( 1 \right) - {\left( { - 3} \right)^2} = - 9 \cr & D < 0{\text{ }},{\text{ There is a saddle point at }}\left( {0,0} \right) \cr & \cr & {\text{For }}{P_1}\left( {3,9} \right) \cr & D = \left( {6\left( 3 \right)} \right)\left( 1 \right) - {\left( { - 3} \right)^2} = 9{\text{ and }}{f_{xx}}\left( {3,9} \right) = 18 \cr & D > 0{\text{ and }}{f_{xx}} > 0,{\text{ There is a relative minimum at }}\left( {3,9} \right) \cr} $$