Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 6

$${\text{Relative minimum at the point }}\left( {5,6, - 3} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = - {x^2} - {y^2} + 10x + 12y - 64 \cr & f\left( {x,y} \right) = - \left( {{x^2} - 10x} \right) - \left( {{y^2} - 12y} \right) - 64 \cr & f\left( {x,y} \right) = - \left( {{x^2} - 10x + 25} \right) - \left( {{y^2} - 12y + 36} \right) - 64 + 25 + 36 \cr & f\left( {x,y} \right) = - {\left( {x - 5} \right)^2} - {\left( {y - 6} \right)^2} - 3 \cr & {\text{Represents an infinite paraboloid with global maximum}} \cr & {\text{at }}\left( {5,6,f\left( {5,6} \right)} \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[ { - {x^2} - {y^2} + 10x + 12y - 64} \right] \cr & {f_x}\left( {x,y} \right) = - 2x + 10 \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - {x^2} - {y^2} + 10x + 12y - 64} \right] \cr & {f_y}\left( {x,y} \right) = - 2y + 12 \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & - 2x + 10 = 0 \cr & x = 5 \cr & - 2y + 12 = 0 \cr & y = 6 \cr & {\text{The critical point is }}\left( {5,6} \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[ { - 2x + 10} \right] = - 2 \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2y + 12} \right] = - 2 \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2x + 10} \right] = 0 \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( {5,6} \right) \cr & d\left( {5,6} \right) = \left( { - 2} \right)\left( { - 2} \right) - {\left[ 0 \right]^2} \cr & d\left( {5,6} \right) = 4 \cr & d > 0,{\text{ and }}{f_{xx}}\left( {5,6} \right) = - 2 < 0 \cr & {\text{then}} \cr & f\left( {x,y} \right){\text{ has a relative maximum at }}\left( {5,6,f\left( {5,6} \right)} \right) \cr & f\left( {5,6} \right) = - {\left( {x - 5} \right)^2} - {\left( {y - 6} \right)^2} - 3 \cr & f\left( {5,6} \right) = - 3 \cr & {\text{Relative maximum at the point }}\left( {5,6, - 3} \right) \cr} $$