Chapter 12 - Functions of Several Veriables - 12.8 Maximum/Minimum Problems - 12.8 Exercises - Page 948: 20

$${\text{Local minimum at }}\left( {\frac{1}{4}, - 2} \right)$$

$$\eqalign{ & f\left( {x,y} \right) = {\left( {4x - 1} \right)^2} + {\left( {2y + 4} \right)^2} + 1 \cr & {\text{Calculate the first partial derivatives}} \cr & {f_x}\left( {x,y} \right) = 2\left( {4x - 1} \right) \cr & {f_x}\left( {x,y} \right) = 8x - 2 \cr & {f_y}\left( {x,y} \right) = 2\left( {2y + 4} \right) \cr & {f_y}\left( {x,y} \right) = 4y + 8 \cr & {\text{Calculate the second partial derivatives and }}{f_{xy}}\left( {x,y} \right) \cr & {f_{xx}}\left( {x,y} \right) = 8 \cr & {f_{yy}}\left( {x,y} \right) = 4 \cr & {f_{xy}}\left( {x,y} \right) = 0 \cr & {\text{Set the first partial derivatives to zero}}{\text{,}} \cr & {f_x}\left( {x,y} \right) = 8x - 2 = 0 \Rightarrow x = \frac{1}{4} \cr & {f_y}\left( {x,y} \right) = 4y + 8 = 0 \Rightarrow y = - 2 \cr & {\text{The critical point is}} \cr & \left( {x,y} \right) = \left( {\frac{1}{4}, - 2} \right) \cr & {\text{Evaluate the Second Derivative Test}} \cr & D\left( {x,y} \right) = {f_x}\left( {x,y} \right){f_y}\left( {x,y} \right) - {\left( {{f_{xy}}\left( {x,y} \right)} \right)^2} \cr & {\text{Evaluate }}D\left( {x,y} \right){\text{ at }}\left( {\frac{1}{4}, - 2} \right) \cr & D\left( {\frac{1}{4}, - 2} \right) = {f_x}\left( {\frac{1}{4}, - 2} \right){f_y}\left( {\frac{1}{4}, - 2} \right) - {\left( {{f_{xy}}\left( {\frac{1}{4}, - 2} \right)} \right)^2} \cr & D\left( {\frac{1}{4}, - 2} \right) = \left( 8 \right)\left( 4 \right) - {\left( 0 \right)^2} \cr & D\left( {\frac{1}{4}, - 2} \right) = 32 \cr & \cr & {f_{xx}}\left( {\frac{1}{4}, - 2} \right) = 8 > 0 \cr & D > 0{\text{ and }}{f_{xx}}\left( {\frac{1}{4}, - 2} \right) > 0 \Rightarrow {\text{Local minimum at }}\left( {\frac{1}{4}, - 2} \right) \cr & f\left( {x,y} \right) = {\left( {4x - 1} \right)^2} + {\left( {2y + 4} \right)^2} + 1 \cr & f\left( {\frac{1}{4}, - 2} \right) = 1 \cr} $$