Question

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

Answer 1

Let
`P=(x,y,z)` be an arbitrary point on the surface. The distance between
`P` and the given point
`(7,11,0)` is given by the function


`d(x,y,z)=√((x-7)^2+(y-11)^2+z^2)`

Note that
`f(x)` and
`f(x)^2` attain their extrema, if they have any, at the same values of
`x`. This allows us to consider the modified distance function,


`d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2`

So now you're minimizing
`d^*(x,y,z)` subject to the constraint
`z^2=xy`. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be


`\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)`

which has partial derivatives


`\begin{cases}(\mathrm d\mathcal L)/(\mathrm dx)=2(x-7)-\lambda y\\\\(\mathrm d\mathcal L)/(\mathrm dy)=2(y-11)-\lambda x\\\\(\mathrm d\mathcal L)/(\mathrm dz)=2z+2\lambda z\\\\(\mathrm d\mathcal L)/(\mathrm d\lambda)=z^2-xy\end{cases}`

Setting all four equation equal to 0, you find from the third equation that either
`z=0` or
`\lambda=-1`. In the first case, you arrive at a possible critical point of
`(0,0,0)`. In the second, plugging
`\lambda=-1` into the first two equations gives


`\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10`

and plugging these into the last equation gives


`z^2=20\implies z=\pm√(20)=\pm2\sqrt5`

So you have three potential points to check:
`(0,0,0)`,
`(2,10,2\sqrt5)`, and
`(2,10,-2\sqrt5)`. Evaluating either distance function (I use
`d^*`), you find that


`d^*(0,0,0)=170`

`d^*(2,10,2\sqrt5)=46`

`d^*(2,10,-2\sqrt5)=46`

So the two points on the surface
`z^2=xy` closest to the point
`(7,11,0)` are
`(2,10,\pm2\sqrt5)`.