1. We are asked to find the point on the surface $z = xy + 1$ that is nearest to the origin $(0,0,0)$. 2. The distance $D$ from any point $(x,y,z)$ to the origin is given by the formula: $$D = \sqrt{x^2 + y^2 + z^2}$$ To minimize the distance, it is easier to minimize the square of the distance: $$D^2 = x^2 + y^2 + z^2$$ 3. Since the point lies on the surface $z = xy + 1$, substitute $z$: $$D^2 = x^2 + y^2 + (xy + 1)^2$$ 4. Expand the square: $$D^2 = x^2 + y^2 + x^2 y^2 + 2xy + 1$$ 5. To find the minimum, take partial derivatives with respect to $x$ and $y$ and set them to zero: $$\frac{\partial D^2}{\partial x} = 2x + 2xy^2 + 2y = 0$$ $$\frac{\partial D^2}{\partial y} = 2y + 2x^2 y + 2x = 0$$ 6. Simplify the system: $$2x + 2xy^2 + 2y = 0 \implies x + xy^2 + y = 0$$ $$2y + 2x^2 y + 2x = 0 \implies y + x^2 y + x = 0$$ 7. Rewrite: $$x(1 + y^2) + y = 0$$ $$y(1 + x^2) + x = 0$$ 8. From the first equation: $$x(1 + y^2) = -y$$ From the second: $$y(1 + x^2) = -x$$ 9. Multiply both sides of the first by $y$ and the second by $x$: $$xy(1 + y^2) = -y^2$$ $$xy(1 + x^2) = -x^2$$ 10. Subtract the two equations: $$xy(1 + y^2) - xy(1 + x^2) = -y^2 + x^2$$ $$xy(y^2 - x^2) = x^2 - y^2$$ 11. Factor the right side: $$xy(y^2 - x^2) = -(y^2 - x^2)$$ 12. If $y^2 \neq x^2$, divide both sides by $(y^2 - x^2)$: $$xy = -1$$ 13. Substitute $xy = -1$ into the surface equation: $$z = xy + 1 = -1 + 1 = 0$$ 14. Now, from the first equation: $$x(1 + y^2) + y = 0$$ Substitute $y = -\frac{1}{x}$ (from $xy = -1$): $$x\left(1 + \frac{1}{x^2}\right) - \frac{1}{x} = 0$$ $$x\left(\frac{x^2 + 1}{x^2}\right) - \frac{1}{x} = 0$$ $$\frac{x(x^2 + 1)}{x^2} - \frac{1}{x} = 0$$ $$\frac{x^3 + x}{x^2} - \frac{1}{x} = 0$$ $$\frac{x^3 + x}{x^2} = \frac{1}{x}$$ 15. Multiply both sides by $x^2$: $$x^3 + x = x$$ $$x^3 = 0$$ $$x = 0$$ 16. If $x=0$, then from $xy = -1$, $0 \cdot y = -1$ which is impossible. 17. Therefore, $y^2 = x^2$ must hold, so $y = \pm x$. 18. Substitute $y = x$ into the first equation: $$x(1 + x^2) + x = 0$$ $$x + x^3 + x = 0$$ $$2x + x^3 = 0$$ $$x(2 + x^2) = 0$$ 19. Solutions: $$x = 0$$ or $$2 + x^2 = 0$$ (no real solution) 20. If $x=0$, then $y=0$, and $z = xy + 1 = 1$. 21. Distance squared at $(0,0,1)$: $$D^2 = 0 + 0 + 1^2 = 1$$ 22. Substitute $y = -x$ into the first equation: $$x(1 + (-x)^2) - x = 0$$ $$x(1 + x^2) - x = 0$$ $$x + x^3 - x = 0$$ $$x^3 = 0$$ $$x=0$$ 23. Then $y = -0 = 0$, and $z = 0 + 1 = 1$ again. 24. So the only candidate point is $(0,0,1)$. 25. Check the distance to origin: $$D = \sqrt{0^2 + 0^2 + 1^2} = 1$$ 26. Conclusion: The point on the surface $z = xy + 1$ nearest to the origin is $(0,0,1)$ with minimum distance 1.