Answer
$$x = 36{\text{ and }}y = 3\sqrt 2 $$
Work Step by Step
$$\eqalign{
& {\text{We have that:}} \cr
& {x^2} + y = 54{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{The product of the numbers}} \cr
& P = xy{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Solve equation }}\left( {\bf{1}} \right){\text{ for }}y \cr
& y = 54 - {x^2} \cr
& {\text{Substitute }}54 - {x^2}{\text{ into equation }}\left( {\bf{2}} \right) \cr
& P = x\left( {54 - {x^2}} \right) \cr
& P = 54x - {x^3} \cr
& {\text{Differentiate}} \cr
& \frac{{dP}}{{dx}} = 54 - 3{x^2} \cr
& {\text{Find the critical points by solving }}\frac{{dP}}{{dx}} = 0 \cr
& 3{x^2} = 54 \cr
& {x^2} = 18 \cr
& x = \pm \sqrt {18} \cr
& x = \pm 3\sqrt 2 \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{{d^2}P}}{{d{x^2}}} = - 6x \cr
& {\text{By the second derivative test:}} \cr
& {\left. {\frac{{{d^2}P}}{{d{x^2}}}} \right|_{x = 3\sqrt 2 }} = - 6\left( {3\sqrt 2 } \right) < 0{\text{ Relative maximum}} \cr
& {\left. {\frac{{{d^2}S}}{{d{x^2}}}} \right|_{x = - 3\sqrt 2 }} = - 6\left( { - 3\sqrt 2 } \right) > 0{\text{ Relative minimum}} \cr
& {\text{Taking }}x = 3\sqrt 2 \cr
& y = 54 - {x^2} \to y = 54 - {\left( {3\sqrt 2 } \right)^2} = 36 \cr
& {\text{Therefore, the numbers are: }}x = 36{\text{ and }}y = 3\sqrt 2 \cr} $$
