Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

Let be an arbitrary point on the surface. The distance between and the given point is given by the function



Note that and attain their extrema, if they have any, at the same values of . This allows us to consider the modified distance function,



So now you're minimizing subject to the constraint . This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be



which has partial derivatives



Setting all four equation equal to 0, you find from the third equation that either or . In the first case, you arrive at a possible critical point of . In the second, plugging into the first two equations gives



and plugging these into the last equation gives



So you have three potential points to check: , , and . Evaluating either distance function (I use ), you find that





So the two points on the surface closest to the point are .

< i think this is the answer unless i did my math wrong