Answer
$$\eqalign{
& {\text{max}} = 16{\text{ at }}x = 4 \cr
& {\text{min}} = 0{\text{ at }}x = 0 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 8x - {x^2};\,\,\left[ {0,6} \right] \cr
& f\left( x \right){\text{ is }}\,{\text{continuous }}\,{\text{and}}\,{\text{differentiable everywhere}}{\text{, the }} \cr
& {\text{absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }}\,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of }} \cr
& {\text{the interval or at solutions to the equation }}f'\left( x \right) = 0{\text{ in}} \cr
& {\text{the open interval }}\left( {0,6} \right){\text{.}} \cr
& \cr
& {\text{Differentiating }}f\left( x \right) \cr
& f'\left( x \right) = 8 - 2x \cr
& {\text{setting }}f'\left( x \right) = 0 \cr
& 8 - 2x = 0 \cr
& - 2x = - 8 \cr
& x = 4 \cr
& {\text{Thus}}{\text{, there is a stationary point at }}x = 4. \cr
& {\text{Evaluating }}f\left( x \right){\text{ at the endpoints}}{\text{, }}x = 0,{\text{ }}x = 6{\text{ and }}x = 4 \cr
& f\left( 0 \right) = 8\left( 0 \right) - {\left( 0 \right)^2} = 0 \cr
& f\left( 4 \right) = 8\left( 4 \right) - {\left( 4 \right)^2} = 16 \cr
& f\left( 6 \right) = 8\left( 6 \right) - {\left( 6 \right)^2} = 12 \cr
& {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr
& {\text{the interval }}\left[ {0,6} \right]{\text{ is 0 at }}x = 0.{\text{ And the absolute maximum is}} \cr
& {\text{16 at }}x = 4\,\,\left( {{\text{Stationary point}}} \right). \cr
& \cr
& {\text{max}} = 16{\text{ at }}x = 4 \cr
& {\text{min}} = 0{\text{ at }}x = 0 \cr} $$
