Chapter 15 - Section 15.3 - Maxima and Minima - Exercises - Page 1117: 15

$${\text{Saddle point at }}\left( {0,0,0} \right)$$

$$\eqalign{ & k\left( {x,y} \right) = {x^2} - 3xy + {y^2} \cr & {\text{Calculate the first partial derivatives of }}k\left( {x,y} \right) \cr & {k_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} - 3xy + {y^2}} \right] \cr & {k_x}\left( {x,y} \right) = 2x - 3y \cr & and \cr & {k_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - 3xy + {y^2}} \right] \cr & {k_y}\left( {x,y} \right) = - 3x + 2y \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {k_x}\left( {x,y} \right) = 0 \cr & 2x - 3y = 0{\text{ }}\left( {\bf{1}} \right) \cr & {k_y}\left( {x,y} \right) = 0 \cr & - 3x + 2y = 0{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solving the system of linear equations we obtain}} \cr & x = 0{\text{ and }}y = 0 \cr & {\text{The critical point is }}\left( {0,0} \right) \cr & {\text{Find the second partial derivatives of }}k\left( {x,y} \right){\text{ and }}{k_{xy}}\left( {x,y} \right) \cr & {k_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x - 3y} \right] = 2 \cr & {k_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 3x + 2y} \right] = 2 \cr & {k_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x - 3y} \right] = - 3 \cr & {\text{By the second partials test}} \cr & d\left( {x,y} \right) = {k_{xx}}\left( {a,b} \right){k_{yy}}\left( {a,b} \right) - {\left[ {{k_{xy}}\left( {x,y} \right)} \right]^2} \cr & {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {0,0} \right) \cr & d\left( {0,0} \right) = {k_{xx}}\left( {0,0} \right){k_{yy}}\left( {0,0} \right) - {\left[ {{k_{xy}}\left( {0,0} \right)} \right]^2} \cr & d\left( {0,0} \right) = \left( 2 \right)\left( 2 \right) - {\left[ { - 3} \right]^2} \cr & d\left( {0,0} \right) = - 5 \cr & d < 0,{\text{ then}} \cr & k\left( {x,y} \right){\text{ has a saddle point at }}\left( {0,0,k\left( {0,0} \right)} \right) \cr & k\left( {x,y} \right) = {x^2} - 3xy + {y^2} \cr & k\left( {x,y} \right) = {\left( 0 \right)^2} - 3\left( 0 \right)\left( 0 \right) + {\left( 0 \right)^2} \cr & {\text{Saddle point at }}\left( {0,0,0} \right) \cr} $$