Answer
$x = 1{\text{ and }}y = 1$
Work Step by Step
$$\eqalign{
& {\text{Let }}x{\text{ and }}y{\text{ the numbers}}{\text{,}} \cr
& \left( {{\text{The second number is reciprocal to the first number}}} \right),{\text{ then}} \cr
& {\text{We have that:}} \cr
& y = \frac{1}{x}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{The sum of the numbers is a minimum }} \cr
& S = x + y{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Substitute }}\frac{1}{x}{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& S = x + \frac{1}{x} \cr
& {\text{Differentiate}} \cr
& \frac{{dS}}{{dx}} = 1 - \frac{1}{{{x^2}}} \cr
& {\text{Find the critical points by solving }}\frac{{dS}}{{dx}} = 0 \cr
& 1 - \frac{1}{{{x^2}}} = 0 \cr
& {x^2} = 1 \cr
& x = \pm 1 \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{{d^2}S}}{{d{x^2}}} = \frac{2}{{{x^3}}} \cr
& {\text{By the second derivative test:}} \cr
& {\left. {\frac{{{d^2}S}}{{d{x^2}}}} \right|_{x = - 1}} = \frac{2}{{{{\left( { - 1} \right)}^3}}} = - 2 < 0{\text{ Relative maximum}} \cr
& {\left. {\frac{{{d^2}S}}{{d{x^2}}}} \right|_{x = 1}} = \frac{2}{{{{\left( 1 \right)}^3}}} = 2 > 0{\text{ Relative minimum}} \cr
& {\text{Taking }}x = 1{\text{ }}\left( {{\text{Because the sum is minimum}}} \right) \cr
& y = \frac{1}{x} \to y = \frac{1}{1} = 1 \cr
& {\text{Therefore}}{\text{, the numbers are: }}x = 1{\text{ and }}y = 1 \cr} $$
