I got a sphere with $x^2+y^2+z^2=1$ and a plane with $x+y+z=1$. I should minimize and maximize the distance from the intersection to (0,0,0).
First thing that threw me off, is that $x^2+y^2+z^2=1$ is centered around $(0,0,0)$. So, every point that is on the sphere, and therefore every point that is on the intersection, has the same distance to $(0,0,0)$, right? Not sure what to minimize?
Since we are so supposed to use Langrage Multipliers, I tried
$$
f:x^2+y^2+z^2=1$$
$$g:x+y+z-1=0
$$
$$
\nabla_f=\nabla_g \lambda,
$$
and got $$\left\{\begin{array}{l}2x=\lambda \\2y=\lambda \\2z=\lambda \end{array}\right.$$
Together with g that gave me $x=y=z=1/3$.
Geometrically I have $(1,0,0),(0,1,0),(0,0,1)$ on my intersection, and I concluded that the intersection is the circle with the center $(1/3,1/3,1/3)$ and radius ${\sqrt2/\sqrt3}$.
Why did the Lagrange Multplier Method give me the center of the circle? Shouldn't it give me the circle itself? I am not sure what I did wrong. Also: Any guesses what one could maximize/minimize here?
Every help is much appreciated!
Best regards!
Best Answer
The problem makes no sense for the reason that you have explained: all points of that line are at the same distance from $(0,0,0)$.
Besides, what you did is not the right approach. Suppose that you want to find the minimum and the maximum of a function $\varphi$ restricted to$$\{(x,y,z)\in\Bbb R^3\mid x^2+y^2+z^2=0\wedge x+y+z=1\}.$$In order to do that, you define $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)=x+y+z$, and then you solve the system$$\left\{\begin{array}{l}\nabla\varphi=\lambda\nabla f+\mu\nabla g\\f(x,y,z)=1\\g(x,y,z)=1\end{array}\right.\iff\left\{\begin{array}{l}\varphi_x=2x\lambda+\mu\\\varphi_y=2y\lambda+\mu\\\varphi_z=2z\lambda+\mu\\x^2+y^2+z^2=1\\x+y+z=1.\end{array}\right.$$
For instance, if you want to find the points of that line which minimize and maximize the distance to $(2,0,0)$, then you can take $\varphi(x,y,z)=(x-2)^2+y^2+z^2$, and then you will get the points $(1,0,0)$ ($\longleftrightarrow$ minimal distance) and $\left(-\frac13,\frac23,\frac23\right)$ ($\longleftrightarrow$ maximal distance).