Answer
$$x = 3{\text{ and }}y = \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}A{\text{ be the area to be minimized}} \cr
& A = xy{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{From the given image in the textbook we have that }}y = \frac{{6 - x}}{2} \cr
& {\text{Substituting }}\frac{{6 - x}}{2}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& A = x\left( {\frac{{6 - x}}{2}} \right) \cr
& A = 3x - \frac{{{x^2}}}{2} \cr
& {\text{Differentiate}} \cr
& \frac{{dA}}{{dx}} = 3 - x \cr
& {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr
& 3 - x = 0 \cr
& x = 3 \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {3 - x} \right] \cr
& \frac{{{d^2}A}}{{d{x^2}}} = - 1 \cr
& {\text{By the second derivative test:}} \cr
& {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = 3}} = - < 0{\text{ Relative maximum}} \cr
& {\text{Calculate }}y \cr
& y = \frac{{6 - x}}{2} \to y = \frac{{6 - 3}}{2} = \frac{3}{2} \cr
& {\text{Therefore, the dimensions are:}} \cr
& x = 3{\text{ and }}y = \frac{3}{2} \cr} $$
