Answer
\[x=5\sqrt{2}\text{ and }y=5\sqrt{2}\]
Work Step by Step
\[\begin{align}
& \text{Let }x\text{ and }y\text{ the numbers, } \\
& \text{The product of the numbers is 50, then} \\
& xy=50\text{ }\left( \mathbf{1} \right) \\
& \text{In this case},\text{ the objective function is the sum }x\text{ and }y \\
& S=x+y\text{ }\left( \mathbf{2} \right) \\
& \text{The domain is }x\ge 0\text{ and }y\ge 0\text{ because are two positive }R \\
& \text{Solve the equation }\left( \mathbf{1} \right)\text{ for }y \\
& y=\frac{50}{x} \\
& \text{Substitute }\frac{50}{x}\text{ for }y\text{ into the equation }\left( \mathbf{2} \right) \\
& S=x+\frac{50}{x} \\
& \text{Differentiate }S\text{ with respect to }x \\
& \frac{dS}{dx}=1-\frac{50}{{{x}^{2}}} \\
& \text{Find the critical points by solving }\frac{dS}{dx}=0 \\
& 1-\frac{50}{{{x}^{2}}}=0 \\
& {{x}^{2}}=50 \\
& x=\pm \sqrt{50} \\
& x=\pm 5\sqrt{2} \\
& x\text{ is positive, then} \\
& x=5\sqrt{2} \\
& \text{Calculating the second derivative of the function} \\
& \frac{{{d}^{2}}S}{d{{x}^{2}}}=\frac{d}{dx}\left[ 1-\frac{50}{{{x}^{2}}} \right] \\
& \frac{{{d}^{2}}S}{d{{x}^{2}}}=0-50\left( -2{{x}^{-3}} \right) \\
& \frac{{{d}^{2}}S}{d{{x}^{2}}}=\frac{100}{{{x}^{3}}} \\
& \text{At }x=5\sqrt{2}\Rightarrow \frac{{{d}^{2}}S}{d{{x}^{2}}}\text{is positive, then there is a local minumum} \\
& \text{Now we can calculate }y \\
& y=\frac{50}{x} \\
& y=\frac{50}{5\sqrt{2}} \\
& y=5\sqrt{2} \\
& \text{Therefore, the numbers are }x=5\sqrt{2}\text{ and }y=5\sqrt{2} \\
\end{align}\]
