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

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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).
User Seybo Glaux
by
8.5k points

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