Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 16

$${\text{The function }}h\left( {x,y} \right){\text{ has a relative minimum at }}\left( {0,0,2} \right)$$

$$\eqalign{ & h\left( {x,y} \right) = {\left( {{x^2} + {y^2}} \right)^{1/3}} + 2 \cr & {\text{Calculate the first partial derivatives of }}h\left( {x,y} \right) \cr & {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\left( {{x^2} + {y^2}} \right)}^{1/3}} + 2} \right] \cr & {h_x}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^2} + {y^2}} \right)^{ - 2/3}}\left( {2x} \right) \cr & {h_x}\left( {x,y} \right) = x{\left( {{x^2} + {y^2}} \right)^{ - 2/3}} \cr & and \cr & {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\left( {{x^2} + {y^2}} \right)}^{1/3}} + 2} \right] \cr & {h_y}\left( {x,y} \right) = y{\left( {{x^2} + {y^2}} \right)^{ - 2/3}} \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {h_x}\left( {x,y} \right) = 0,{\text{ }}{h_y}\left( {x,y} \right) = 0 \cr & x{\left( {{x^2} + {y^2}} \right)^{ - 2/3}} = 0,{\text{ }}y{\left( {{x^2} + {y^2}} \right)^{ - 2/3}} = 0 \cr & {\text{The critical point is }}\left( {0,0} \right) \cr & {\text{Solving the system of equations, we obtain}} \cr & x = 0,{\text{ }}y = 0 \cr & {\text{The critical point is }}\left( {0,0} \right) \cr & {\text{The function }}{\left( {{x^2} + {y^2}} \right)^{1/3}} + 2{\text{ is always positive for all real }} \cr & {\text{numbers}}{\text{. Because the critical point is }}\left( {0,0} \right){\text{ we can conclude}} \cr & {\text{that is a relative minumum. Thus, we find: }} \cr & h\left( {x,y} \right) = {\left( {{x^2} + {y^2}} \right)^{1/3}} + 2 \cr & h\left( {0,0} \right) = {\left( {{0^2} + {0^2}} \right)^{1/3}} + 2 \cr & h\left( {0,0} \right) = 2 \cr & {\text{The function }}h\left( {x,y} \right){\text{ has a relative minimum at }}\left( {0,0,2} \right) \cr} $$