Chapter 13 - Functions of Several Variables - 13.8 Exercises - Page 942: 11

$${\text{Relative maximum at the point }}\left( {\frac{1}{2}, - 1,\frac{{31}}{4}} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = - 3{x^2} - 2{y^2} + 3x - 4y + 5 \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[ { - 3{x^2} - 2{y^2} + 3x - 4y + 5} \right] \cr & {f_x}\left( {x,y} \right) = - 6x + 3 \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 3{x^2} - 2{y^2} + 3x - 4y + 5} \right] \cr & {f_y}\left( {x,y} \right) = - 4y - 4 \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 & - 6x + 3 = 0,{\text{ }} - 4y - 4 = 0 \cr & x = \frac{1}{2},{\text{ }}y = - 1 \cr & {\text{The critical point is }}\left( {\frac{1}{2}, - 1} \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[ { - 6x + 3} \right] = - 6 \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 4y - 4} \right] = - 2 \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 6x + 3} \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( {\frac{1}{2}, - 1} \right) \cr & d\left( {\frac{1}{2}, - 1} \right) = {f_{xx}}\left( {\frac{1}{2}, - 1} \right){f_{yy}}\left( {\frac{1}{2}, - 1} \right) - {\left[ {{f_{xy}}\left( {\frac{1}{2}, - 1} \right)} \right]^2} \cr & d\left( {\frac{1}{2}, - 1} \right) = \left( { - 6} \right)\left( { - 2} \right) - {\left[ 0 \right]^2} \cr & d\left( {\frac{1}{2}, - 1} \right) = 12 \cr & d > 0,{\text{ and }}{f_{xx}}\left( {\frac{1}{2}, - 1} \right) < 0 \cr & f\left( {x,y} \right){\text{ has a relative maximum at }}\left( {\frac{1}{2}, - 1,f\left( {\frac{1}{2}, - 1} \right)} \right) \cr & f\left( {x,y} \right) = - 3{x^2} - 2{y^2} + 3x - 4y + 5 \cr & f\left( {\frac{1}{2}, - 1} \right) = - 3{\left( {\frac{1}{2}} \right)^2} - 2{\left( { - 1} \right)^2} + 3\left( {\frac{1}{2}} \right) - 4\left( { - 1} \right) + 5 = \frac{{31}}{4} \cr & {\text{Relative maximum at the point }}\left( {\frac{1}{2}, - 1,\frac{{31}}{4}} \right) \cr} $$