Answer
$${\text{Local minimum at }}\left( {1,2} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^4} + {y^4} - 4x - 32y + 10 \cr
& {\text{Calculate the first partial derivatives}} \cr
& {f_x}\left( {x,y} \right) = 4{x^3} - 4 \cr
& {f_y}\left( {x,y} \right) = 4{y^3} - 32 \cr
& {\text{Calculate the second partial derivatives and }}{f_{xy}}\left( {x,y} \right) \cr
& {f_{xx}}\left( {x,y} \right) = 12{x^2} \cr
& {f_{yy}}\left( {x,y} \right) = 12{y^2} \cr
& {f_{xy}}\left( {x,y} \right) = 0 \cr
& {\text{Set the first partial derivatives to zero}}{\text{,}} \cr
& {f_x}\left( {x,y} \right) = 4{x^3} - 4 \Rightarrow x = 1 \cr
& {f_y}\left( {x,y} \right) = 4{y^3} - 32 \Rightarrow y = 2 \cr
& {\text{The critical point is}} \cr
& \left( {x,y} \right) = \left( {1,2} \right) \cr
& {\text{Evaluate the Second Derivative Test}} \cr
& D\left( {x,y} \right) = {f_x}\left( {x,y} \right){f_y}\left( {x,y} \right) - {\left( {{f_{xy}}\left( {x,y} \right)} \right)^2} \cr
& {\text{Evaluate }}D\left( {x,y} \right){\text{ at }}\left( {1,2} \right) \cr
& D\left( {1,2} \right) = {f_x}\left( {1,2} \right){f_y}\left( {1,2} \right) - {\left( {{f_{xy}}\left( {1,2} \right)} \right)^2} \cr
& D\left( {1,2} \right) = \left( {12{{\left( 1 \right)}^2}} \right)\left( {12{{\left( 2 \right)}^2}} \right) - {\left( 0 \right)^2} \cr
& D\left( {1,2} \right) = 576 \cr
& \cr
& {f_{xx}}\left( {1,2} \right) = 12{\left( 1 \right)^2} = 12 \cr
& D > 0{\text{ and }}{f_{xx}}\left( {1,2} \right) > 0 \Rightarrow {\text{Local minimum at }}\left( {1,2} \right) \cr} $$
