Chapter 13 - Functions of Several Variables - Review Exercises - Page 962: 65

$${\text{Relative minimum at the point }}\left( {1,1,3} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = xy + \frac{1}{x} + \frac{1}{y} \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[ {xy + \frac{1}{x} + \frac{1}{y}} \right] \cr & {f_x}\left( {x,y} \right) = y - \frac{1}{{{x^2}}} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {xy + \frac{1}{x} + \frac{1}{y}} \right] \cr & {f_y}\left( {x,y} \right) = x - \frac{1}{{{y^2}}} \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {f_x}\left( {x,y} \right) = y - \frac{1}{{{x^2}}} = 0,{\text{ }}{f_y}\left( {x,y} \right) = y - \frac{1}{{{x^2}}} = 0 \cr & {\text{By symmetry }}x = y,{\text{ then}} \cr & x - \frac{1}{{{x^2}}} = 0,{\text{ }}y - \frac{1}{{{y^2}}} = 0 \cr & x = 1,{\text{ }}y = 1 \cr & {\text{The critical point is }}\left( {1,1} \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[ {y - \frac{1}{{{x^2}}}} \right] = \frac{2}{{{x^3}}} \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x - \frac{1}{{{y^2}}}} \right] = \frac{2}{{{y^3}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {y - \frac{1}{{{x^2}}}} \right] = 1 \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( { - 1,1} \right) \cr & d\left( {1,1} \right) = {f_{xx}}\left( {1,1} \right){f_{yy}}\left( {1,1} \right) - {\left[ {{f_{xy}}\left( {1,1} \right)} \right]^2} \cr & d\left( {1,1} \right) = \left( {\frac{2}{{{{\left( 1 \right)}^3}}}} \right)\left( {\frac{2}{{{{\left( 1 \right)}^3}}}} \right) - {\left[ 1 \right]^2} \cr & d\left( {1,1} \right) = 3 \cr & d > 0,{\text{ and }}{f_{xx}}\left( { - 1,1} \right) = 1 > 0 \cr & {\text{Then}} \cr & f\left( {x,y} \right){\text{ has a relative minimum at }}\left( {1,1,f\left( {1,1} \right)} \right) \cr & f\left( {x,y} \right) = xy + \frac{1}{x} + \frac{1}{y} \cr & f\left( {1,1} \right) = \left( 1 \right)\left( 1 \right) + \frac{1}{1} + \frac{1}{1} \cr & f\left( {1,1} \right) = 3 \cr & {\text{Relative minimum at the point }}\left( {1,1,3} \right) \cr} $$