Chapter 13 - Functions of Several Variables - Review Exercises - Page 962: 66

$${\text{Relative maximum at the point }}\left( {\frac{3}{2},3,16} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = - 8{x^2} + 4xy - {y^2} + 12x + 7 \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[ { - 8{x^2} + 4xy - {y^2} + 12x + 7} \right] \cr & {f_x}\left( {x,y} \right) = - 16x + 4y + 12 \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 8{x^2} + 4xy - {y^2} + 12x + 7} \right] \cr & {f_y}\left( {x,y} \right) = 4x - 2y \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 & - 16x + 4y + 12 = 0,{\text{ }}4x - 2y = 0 \cr & {\text{Solving the system of equations, we obtain}} \cr & x = \frac{3}{2},{\text{ }}y = 3 \cr & {\text{The critical point is }}\left( {\frac{3}{2},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[ { - 16x + 4y + 12} \right] = - 16 \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4x - 2y} \right] = - 2 \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 16x + 4y + 12} \right] = 4 \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( {\frac{3}{2},3} \right) \cr & d\left( {\frac{3}{2},3} \right) = {f_{xx}}\left( {\frac{3}{2},3} \right){f_{yy}}\left( {\frac{3}{2},3} \right) - {\left[ {{f_{xy}}\left( {\frac{3}{2},3} \right)} \right]^2} \cr & d\left( {\frac{3}{2},3} \right) = \left( { - 16} \right)\left( { - 2} \right) - {\left[ 4 \right]^2} \cr & d\left( {\frac{3}{2},3} \right) = 16 \cr & d > 0,{\text{ and }}{f_{xx}}\left( {\frac{3}{2},3} \right) = - 16 < 0 \cr & {\text{then}} \cr & f\left( {x,y} \right){\text{ has a relative maximum at }}\left( {\frac{3}{2},3,f\left( {\frac{3}{2},3} \right)} \right) \cr & f\left( {x,y} \right) = - 8{x^2} + 4xy - {y^2} + 12x + 7 \cr & f\left( {\frac{3}{2},3} \right) = - 8{\left( {\frac{3}{2}} \right)^2} + 4\left( {\frac{3}{2}} \right)\left( 3 \right) - {\left( 3 \right)^2} + 12\left( {\frac{3}{2}} \right) + 7 \cr & f\left( {\frac{3}{2},3} \right) = 16 \cr & {\text{Relative maximum at the point }}\left( {\frac{3}{2},3,16} \right) \cr} $$