We are tasked with finding the point(s) on the surface $z^2 = xy + 1$ that are closest to the point $(10, 8, 0)$.

Problem breakdown:

1. The surface is given by the equation $z^2 = xy + 1$, which describes a relation between the coordinates $(x, y, z)$.
2. We are trying to minimize the distance between any point $(x, y, z)$ on this surface and the point $(10, 8, 0)$.

Step 1: Distance Formula

The distance $D$ between a point $(x, y, z)$ on the surface and $(10, 8, 0)$ is given by the Euclidean distance formula: $D = \sqrt{(x - 10)^2 + (y - 8)^2 + z^2}$ Since minimizing $D$ is equivalent to minimizing $D^2$, we can work with: $D^2 = (x - 10)^2 + (y - 8)^2 + z^2$ But from the surface equation $z^2 = xy + 1$, we can substitute $z^2$ as $xy + 1$. So the expression for $D^2$ becomes: $D^2 = (x - 10)^2 + (y - 8)^2 + (xy + 1)$ Thus, we now have the objective function: $f(x, y) = (x - 10)^2 + (y - 8)^2 + xy + 1$ which we aim to minimize.

Step 2: Gradient and Critical Points

To minimize the function $f(x, y)$, we need to find the partial derivatives with respect to $x$ and $y$, set them to zero, and solve for $x$ and $y$.

Let me compute that.The critical point that minimizes the distance function is $(x, y) = (8, 4)$.

Step 3: Find the corresponding $z$-value

To find the $z$-coordinate corresponding to this point on the surface, we substitute $x = 8$ and $y = 4$ into the surface equation $z^2 = xy + 1$: $z^2 = 8 \cdot 4 + 1 = 33$ Thus, $z = \pm \sqrt{33}$.

Final Answer:

The points on the surface that are closest to $(10, 8, 0)$ are: $(8, 4, \sqrt{33}) \quad \text{and} \quad (8, 4, -\sqrt{33})$

Do you want further details on any step, or any clarification?

Here are 5 related questions:

1. How can we generalize this method for other surfaces?
2. What are the second-order conditions for confirming a local minimum?
3. What happens if the surface is defined by a different equation, such as a plane?
4. How does this method work in higher dimensions (e.g., in 4D space)?
5. Can the distance minimization problem be solved using Lagrange multipliers?

Tip: Minimizing the square of a distance often simplifies computations without affecting the result, as the square preserves the relative ordering of distances.