Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 1

$${\text{Relative minimum at the point }}\left( {1,3,0} \right)$$

$$\eqalign{ & g\left( {x,y} \right) = {\left( {x - 1} \right)^2} + {\left( {y - 3} \right)^2} \cr & {\text{Let }}z = g\left( {x,y} \right),{\text{ the equation is in the form}}{\text{.}} \cr & z = {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} \cr & {\text{Represents an infinite paraboloid with global minimum }} \cr & {\text{at }}\left( {a,b} \right).{\text{ Then,}} \cr & {\text{Calculate the first partial derivatives of }}g\left( {x,y} \right) \cr & {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}} \right] \cr & {g_x}\left( {x,y} \right) = 2\left( {x - 1} \right) \cr & and \cr & {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\left( {x - 1} \right)}^2} + {{\left( {y - 3} \right)}^2}} \right] \cr & {g_y}\left( {x,y} \right) = 2\left( {y - 3} \right) \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {g_x}\left( {x,y} \right) = 0 \cr & 2\left( {x - 1} \right) = 0 \cr & x = 1 \cr & {g_y}\left( {x,y} \right) = 0 \cr & 2\left( {y - 3} \right) = 0 \cr & y = 3 \cr & {\text{The critical point is }}\left( {1,3} \right) \cr & {\text{Find the second partial derivatives of }}g\left( {x,y} \right){\text{ and }}{g_{xy}}\left( {x,y} \right) \cr & {g_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2\left( {x - 1} \right)} \right] = 2 \cr & {g_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2\left( {y - 3} \right)} \right] = 2 \cr & {g_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2\left( {x - 1} \right)} \right] = 0 \cr & {\text{By the second partials test}} \cr & d\left( {x,y} \right) = {g_{xx}}\left( {a,b} \right){g_{yy}}\left( {a,b} \right) - {\left[ {{g_{xy}}\left( {x,y} \right)} \right]^2} \cr & {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {1,3} \right) \cr & d\left( {1,3} \right) = \left( 2 \right)\left( 2 \right) - {\left[ 0 \right]^2} \cr & d\left( {1,3} \right) = 4 \cr & d > 0,{\text{ and }}{g_{xx}}\left( {1,3} \right) = 2 > 0 \cr & {\text{then}} \cr & g\left( {x,y} \right){\text{ has a relative minimum at }}\left( {1,3,g\left( {1,3} \right)} \right) \cr & g\left( {1,3} \right) = {\left( {1 - 1} \right)^2} + {\left( {3 - 3} \right)^2} \cr & g\left( {1,3} \right) = 0 \cr & {\text{Relative minimum at the point }}\left( {1,3,0} \right) \cr} $$