1. **Problem statement:** Find two positive numbers $x$ and $y$ such that their product is 250, i.e., $$xy=250,$$ and the sum $$S = x + 4y$$ is minimized. 2. **Formula and approach:** We want to minimize $$S = x + 4y$$ subject to the constraint $$xy=250.$$ We can express one variable in terms of the other using the constraint: $$x = \frac{250}{y}.$$ Substitute into $S$: $$S(y) = \frac{250}{y} + 4y.$$ 3. **Find critical points:** To minimize $S(y)$ for $y > 0$, differentiate and set derivative to zero: $$\frac{dS}{dy} = -\frac{250}{y^2} + 4 = 0.$$ 4. **Solve for $y$:** $$-\frac{250}{y^2} + 4 = 0 \implies 4 = \frac{250}{y^2} \implies y^2 = \frac{250}{4} = 62.5.$$ So, $$y = \sqrt{62.5} = \frac{5\sqrt{10}}{2}.$$ 5. **Find $x$:** Using $x = \frac{250}{y}$, $$x = \frac{250}{\frac{5\sqrt{10}}{2}} = 250 \times \frac{2}{5\sqrt{10}} = \frac{500}{5\sqrt{10}} = \frac{100}{\sqrt{10}} = 10\sqrt{10}.$$ 6. **Verify minimum:** The second derivative is $$\frac{d^2S}{dy^2} = \frac{500}{y^3} > 0$$ for $y > 0$, confirming a minimum. 7. **Final answer:** The two positive numbers are $$x = 10\sqrt{10}$$ and $$y = \frac{5\sqrt{10}}{2}.$$