calculuslagrange multipliermultivariable-calculussurfaces
I'm attempting to figure this problem out. I would appreciate some guidance on how to get the answer. Thanks.
Consider the surface defined as $S: x^2+y^2+z^2 = 8$. If we have a $P = (0,1,1)$, use Lagrange multipliers to determine the point on $S$ nearest to $P$.
Here is a moderately complicated example, cooked up in such a way that it can be solved explicitly all the way to the end.
Given are the two points $A=(0,-7)$, $B=(-7,0)$, and the circle $\gamma: \>x^2+y^2=4$. Determine two points $P$, $Q\in\gamma$ such that the quantity $$d(P,Q):=|AP|^2+|PQ|^2+|QB|^2$$ becomes maximal, resp., minimal.
You will obtain four conditionally stationary situations $(P_k,Q_k)$: the two extremal situations and two "saddle points". The latter had to be expected, since the surface determined by the conditions is a torus.
A full solution (in German) is given on pp. 212–214 of the following pdf-file. The file contains a full chapter (pages 128–236) of a textbook for engineering students. Scroll down to page 212; there you will see the figure printed above.
You can formulate problem (a) as
\begin{align} \operatorname{maximize} & \quad x \\ \text{subject to} & \quad 4x^2 + 7xy + 8y^2 = 60. \end{align} The optimization variables are $x, y \in \mathbb R$.
You can formulate problem (b) as \begin{align} \operatorname{minimize} & \quad y \\ \text{subject to} & \quad 4x^2 + 7xy + 8y^2 = 60. \end{align} The optimization variables are again $x, y \in \mathbb R$.
Best Answer
It is obvious that the shortest distance between $P$ and $S$ is the distance between $P$ and the center of the sphere, minus the radius, that is: $$ |\sqrt{0^2+1^2+1^2}-\sqrt{8}| = \sqrt{2} $$
Now, if you want the coordinates of the nearest point, you can use Lagrange multipliers indeed: you want to minimize the (squared) distance between $P$ and $S$, given by $$ x^2+(y-1)^2+(z-1)^2 $$ subject to $$ x^2+y^2+z^2=8 $$ The Lagrangian equals $$ \mathcal{L}=x^2+(y-1)^2+(z-1)^2 + \lambda(8-x^2+y^2+z^2) $$ So you need to solve the following system \begin{cases} 2x-2\lambda x =0 \\ 2y-2-2\lambda y = 0 \\ 2z -2 -2\lambda z = 0 \\ x^2+y^2+z^2=8 \end{cases} You should get $$ (x,y,z,\lambda)=(0,2,2,\frac{1}{2}) $$ So the nearest the point has coordinates $(0,2,2)$ and the shortest distance equals $$\sqrt{0^2+(2-1)^2+(2-1)^2}=\sqrt{2}$$