Chapter 15 - Section 15.3 - Maxima and Minima - Exercises - Page 1117: 16

$${\text{Relative minimum at the point }}\left( {0,0,0} \right)$$

$$\eqalign{ & k\left( {x,y} \right) = {x^2} - xy + 2{y^2} \cr & {\text{Calculate the first partial derivatives of }}g\left( {x,y} \right) \cr & {k_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} - xy + 2{y^2}} \right] \cr & {k_x}\left( {x,y} \right) = 2x - y \cr & and \cr & {k_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} - xy + 2{y^2}} \right] \cr & {k_y}\left( {x,y} \right) = - x + 4y \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {k_x}\left( {x,y} \right) = 0 \cr & 2x - y = 0{\text{ }}\left( {\bf{1}} \right) \cr & {k_y}\left( {x,y} \right) = 0 \cr & - x + 4y = 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 - y} \right] = 2 \cr & {k_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - x + 4y} \right] = 4 \cr & {k_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2x - y} \right] = - 1 \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( 4 \right) - {\left[ { - 1} \right]^2} \cr & d\left( {0,0} \right) = 7 \cr & d > 0,{\text{ and }}{k_{xx}}\left( {0,0} \right) = 2 > 0{\text{ then}} \cr & k\left( {x,y} \right){\text{ has a relative minimum at }}\left( {0,0,f\left( {0,0} \right)} \right) \cr & k\left( {0,0} \right) = {\left( 0 \right)^2} - \left( 0 \right)\left( 0 \right) + 2{\left( 0 \right)^2} \cr & k\left( {0,0} \right) = 0 \cr & {\text{Relative minimum at the point }}\left( {0,0,0} \right) \cr} $$