Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 10

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

$$\eqalign{ & h\left( {x,y} \right) = {x^2} - 3xy - {y^2} \cr & {\text{Calculate the first partial derivatives of }}h\left( {x,y} \right) \cr & {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} - 3xy - {y^2}} \right] \cr & {h_x}\left( {x,y} \right) = 2x - 3y \cr & and \cr & {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - 3xy - {y^2}} \right] \cr & {h_y}\left( {x,y} \right) = - 3x - 2y \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {h_x}\left( {x,y} \right) = 0,{\text{ }}{h_y}\left( {x,y} \right) = 0 \cr & 2x - 3y = 0,{\text{ }} - 3x - 2y = 0 \cr & {\text{Solving the system of equations, we obtain}} \cr & x = 0,{\text{ }}y = 0 \cr & {\text{The critical point is }}\left( {0,0} \right) \cr & {\text{Find the second partial derivatives of }}h\left( {x,y} \right){\text{ and }}{h_{xy}}\left( {x,y} \right) \cr & {h_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2x - 3y} \right] = 2 \cr & {h_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 3x - 2y} \right] = - 2 \cr & {h_{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) = {h_{xx}}\left( {a,b} \right){h_{yy}}\left( {a,b} \right) - {\left[ {{h_{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[ 3 \right]^2} \cr & d\left( {\frac{1}{2}, - \frac{1}{2}} \right) = - 13 \cr & d < 0,{\text{ }} \cr & h\left( {x,y} \right){\text{ has a saddle point at }}\left( {0,0,h\left( {0,0} \right)} \right) \cr & h\left( {x,y} \right) = {x^2} - 3xy - {y^2} \cr & h\left( {0,0} \right) = 0 \cr & {\text{Saddle point at }}\left( {0,0,0} \right) \cr} $$