Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 3

$${\text{Relative minimum at the point }}\left( {0,0,1} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = \sqrt {{x^2} + {y^2} + 1} \cr & \sqrt {{x^2} + {y^2} + 1} {\text{ is always positive, so there is a minimum }} \cr & {\text{when }}x = 0{\text{ and }}y = 0. \cr & {\text{Global maximum at }}\left( {0,0,1} \right).{\text{ Then,}} \cr & {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sqrt {{x^2} + {y^2} + 1} } \right] \cr & {f_x}\left( {x,y} \right) = \frac{{2x}}{{2\sqrt {{x^2} + {y^2} + 1} }} \cr & {f_x}\left( {x,y} \right) = \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sqrt {{x^2} + {y^2} + 1} } \right] \cr & {f_y}\left( {x,y} \right) = \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }} \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {f_x}\left( {x,y} \right) = \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }} \cr & \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }} = 0 \cr & x = 0 \cr & {f_y}\left( {x,y} \right) = \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }} \cr & \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }} = 0 \cr & y = 0 \cr & {\text{The critical point is }}\left( {0,0} \right) \cr & {\text{Find the second partial derivatives of }}f\left( {x,y} \right){\text{ and }}{f_{xy}}\left( {x,y} \right) \cr & {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{x}{{\sqrt {{x^2} + {y^2} + 1} }}} \right] = \frac{{{y^2} + 1}}{{{{\left( {{x^2} + {y^2} + 1} \right)}^{3/2}}}} \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right] = \frac{{{x^2} + 1}}{{{{\left( {{x^2} + {y^2} + 1} \right)}^{3/2}}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2\left( {x - 3} \right)} \right] = - \frac{{xy}}{{{{\left( {{x^2} + {y^2} + 1} \right)}^{3/2}}}} \cr & {\text{By the second partials test}} \cr & d\left( {x,y} \right) = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left[ {{f_{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) = \left( 1 \right)\left( 1 \right) - {\left[ 0 \right]^2} \cr & d\left( {0,0} \right) = 1 \cr & d > 0,{\text{ and }}{f_{xx}}\left( {0,0} \right) = 1 > 0 \cr & {\text{then}} \cr & f\left( {x,y} \right){\text{ has a relative minimum at }}\left( {0,0,f\left( {0,0} \right)} \right) \cr & f\left( {0,0} \right) = \sqrt {{0^2} + {0^2} + 1} \cr & f\left( {0,0} \right) = 1 \cr & {\text{Relative minimum at the point }}\left( {0,0,1} \right) \cr} $$