Chapter 13 - Functions of Several Variables - Review Exercises - Page 962: 63

$${\text{Relative minimum at the point }}\left( { - 4,\frac{4}{3}, - 2} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = 2{x^2} + 6xy + 9{y^2} + 8x + 14 \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[ {2{x^2} + 6xy + 9{y^2} + 8x + 14} \right] \cr & {f_x}\left( {x,y} \right) = 4x + 6y + 8 \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2{x^2} + 6xy + 9{y^2} + 8x + 14} \right] \cr & {f_y}\left( {x,y} \right) = 6x + 18y \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {f_x}\left( {x,y} \right) = 0,{\text{ }}{f_y}\left( {x,y} \right) = 0 \cr & 4x + 6y + 8 = 0,{\text{ }}6x + 18y = 0 \cr & {\text{Solving the system of equations, we obtain}} \cr & x = - 4,{\text{ }}y = \frac{4}{3} \cr & {\text{The critical point is }}\left( { - 4,\frac{4}{3}} \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[ {4x + 6y + 8} \right] = 4 \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6x + 18y} \right] = 18 \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4x + 6y + 8} \right] = 6 \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( { - 4,\frac{4}{3}} \right) = {f_{xx}}\left( { - 4,\frac{4}{3}} \right){f_{yy}}\left( { - 4,\frac{4}{3}} \right) - {\left[ {{f_{xy}}\left( { - 4,\frac{4}{3}} \right)} \right]^2} \cr & d\left( { - 4,\frac{4}{3}} \right) = \left( 4 \right)\left( {18} \right) - {\left[ 6 \right]^2} \cr & d\left( { - 4,\frac{4}{3}} \right) = 36 \cr & d > 0,{\text{ and }}{f_{xx}}\left( { - 1,1} \right) = 4 > 0 \cr & {\text{Then}} \cr & f\left( {x,y} \right){\text{ has a relative minimum at }}\left( { - 4,\frac{4}{3},f\left( { - 4,\frac{4}{3}} \right)} \right) \cr & f\left( {x,y} \right) = 2{x^2} + 6xy + 9{y^2} + 8x + 14 \cr & f\left( { - 4,\frac{4}{3}} \right) = 2{\left( { - 4} \right)^2} + 6\left( { - 4} \right)\left( {\frac{4}{3}} \right) + 9{\left( {\frac{4}{3}} \right)^2} + 8\left( { - 4} \right) + 14 \cr & f\left( { - 4,\frac{4}{3}} \right) = - 2 \cr & {\text{Relative minimum at the point }}\left( { - 4,\frac{4}{3}, - 2} \right) \cr} $$