Finding shortest distance between a point and a surface Consider the surface $S$ (in $\mathbb R^3$) given by the equation $z=f(x,y)=\frac32(x^2+y^2)$. How can I find the shortest distance from a point $p=(a,b,c)$ on $S$ to the point $(0,0,1)$.
This is what I have done: Define $d(a,b,c)=a^2+b^2+(c-1)^2$, for all points $p=(a,b,c)\in S$. Then $\sqrt d$ is the distance from $S$ to $(0,0,1)$. I think that the method of Lagrange multipliers is the easiest way to solve my question, but how can I find the Lagrangian function? Or is there an easier way to find the shortest distance?


I think that the method of Lagrange multipliers is the easiest way to
solve my question, but how can I find the Lagrangian function?

As shown by other answers and in note 1 there are easier ways to find the shortest distance, but here is a detailed solution using the method of Lagrange multipliers. You need to find the minimum of the distance function
$$\begin{equation}
d(x,y,z)=\sqrt{x^{2}+y^{2}+(z-1)^{2}}  \tag{1a}
\end{equation}$$
subject to the constraint given by the surface equation $z=f(x,y)=\frac32(x^2+y^2)$
$$\begin{equation}
g(x,y,z)=z-\frac{3}{2}\left( x^{2}+y^{2}\right) =0.  \tag{2}
\end{equation}$$
Since $\sqrt{x^{2}+y^{2}+(z-1)^{2}}$ increases with $x^{2}+y^{2}+(z-1)^{2}$ you can simplify the
computations if you find the minimum of
$$\begin{equation} 
[d(x,y,z)]^2=x^{2}+y^{2}+(z-1)^{2}  \tag{1b}
\end{equation}$$
subject to the same constraint $(2)$. The Lagrangian function is then
defined by
$$\begin{eqnarray}
L\left( x,y,z,\lambda \right)  &=&[d(x,y,z)]^2+\lambda g(x,y,z)  \\
L\left( x,y,z,\lambda \right)  &=&x^{2}+y^{2}+(z-1)^{2}+\lambda \left( z-
\frac{3}{2}\left( x^{2}+y^{2}\right) \right),   \tag{3}
\end{eqnarray}$$
where $\lambda $ is the Lagrange multiplier. By this method you need to
solve the following system
$$\begin{equation}
\left\{ \frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial y}=0,
\frac{\partial L}{\partial z}=0,\frac{\partial L}{\partial \lambda }
=0,\right.   \tag{4}
\end{equation}$$
which results in
$$\begin{eqnarray}
\left\{ 
\begin{array}{c}
2x+3\lambda x=0 \\ 
2y+3\lambda y=0 \\ 
2z-2-\lambda =0 \\ 
-z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0
\end{array}
\right.  &\Leftrightarrow &\left\{ 
\begin{array}{c}
x=0\vee 2+3\lambda =0 \\ 
y=0\vee 2+3\lambda =0 \\ 
2z-2-\lambda =0 \\ 
-z+\frac{3}{2}\left( x^{2}+y^{2}\right) =0
\end{array}
\right.   \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
x=0 \\ 
y=0 \\ 
\lambda =2 \\ 
z=0
\end{array}
\right. \vee \left\{ 
\begin{array}{c}
\lambda =-2/3 \\ 
z=2/3 \\ 
x^{2}+y^{2}=4/9
\end{array}
\right.   \tag{5}
\end{eqnarray}$$
For $x=y=x=0$ we get $d(0,0,0)=1$. And for $x^2+y^2=4/9,z=2/3$ we get the minimum distance subject to the given conditions
$$\begin{equation}
\underset{g(x,y,z)=0}\min d(x,y,z)=\sqrt{\frac{4}{9}+(\frac{2}{3}-1)^{2}}=\frac{1}{3}\sqrt{5}.  \tag{6}
\end{equation}$$
It is attained on the intersection of the surface $z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $ with the vertical cylinder $x^{2}+y^{2}=\frac{4}{9}$ or equivalently with the horizontal plane $z=\frac{2}{3}$.

$$\text{Plane }z=\frac{2}{3} \text{(blue) and surface  }z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $$
Notes.

*

*As the solution depends only on the sum $r^{2}=x^{2}+y^{2}$ we could just find
$$\begin{equation}
\min [d(r)]^2=r^{2}+(\frac{3}{2}r^{2}-1)^{2}  \tag{7}
\end{equation}$$
and then find $d(r)=\sqrt{[d(r)]^2}$ at the minimum.

*The surface $z=\frac{3}{2}\left(
x^{2}+y^{2}\right) $ is a surface of revolution around the $z$ axis.

