boundaries of a hyper surface I have multidimensional surface $f(x_1, x_2, ...., x_n) =c$ in n-dimension. 
 What is the way to figure out what is the minimum value of any one of the dimension (say $x_i$ ) for which the $f(x_1, x_2, ...., x_n) =c$ holds.
To give a specific example:say I have a sphere in 10-dimension with a radius of say 1. Then my function will look like $x_1^2+...x_{10}^2$ and say I choose $x_i=x_5$ then I know the minimal value of $x_5=-1$.
For a arbitrary function what is the way to figure this out. Is this a minimization problem where I compute out $ \min x_i$ subject to the constraint $f(x_1, x_2, ...., x_n) =c$ ?
 A: Your problem can naturally be posed as an optimization problem as follows
$$
\begin{align}
\text{minimize } & x_i\\
\text{subject to }& f(x_1,x_2,\ldots,x_n)=0, 
\end{align}
$$
with optimization variables $x_1,x_2,\ldots,x_n$ (I have moved the constant $c$ to the left-hand-side and incorporated it into the definition of $f$). 
If $f$ is differentiable with respect to $\{x_i\}_{i=1}^n$ you may be able to find the solution by hand, using Lagrange multipliers. In particular, defining the Lagrangian as
$$
\mathcal{L}\triangleq x_i- \lambda(c-f(x_1,x_2,\ldots,x_n)),
$$
where $\lambda \in \mathbb{R}$ is a Lagrange multiplier,
 the optimal solution of the problem (which also provides the minimum $x_i$) is the solution of the system 
$$
\left\{\begin{align}
\frac{\partial \mathcal{L}}{\partial x_j}&= 0,j\in\{1,2,\ldots,n\},\\
f&=0
\end{align}\right\} 
$$
which can be written as
$$
\left\{\begin{align}
1+\lambda \frac{\partial f}{\partial x_i}&=0,\\
\lambda \frac{\partial f}{\partial x_j}&= 0,j\in\{1,2,\ldots,n\} \setminus\ \{i\} \\
f&=0
\end{align}\right\}.
$$
For example, with $f=\sum_{i=1}^n x_i^2 - c$, one can easily verify that the solution of the above system is unique and equals 
$$
x_j=0,\forall j\neq i, \\\lambda = -\frac{1}{2\sqrt{c}},\\ x_i = -\sqrt{c}.
$$
Actually, when $f$ is differentiable, there is a second method to find the minimum $x_i$. In particular, viewing the surface equation $f=0$ as an implicit equation of the (dependent) variable $x_i$ with respect to the (independent) variables $\{x_j\}_{j\neq i}$, one may invoke the implicit function theorem in order to find the gradient of $x_i$ w.r.t. the indepedent variables as
$$
\frac{\partial x_i}{\partial x_j}=-\frac{\partial f /\partial x_j}{\partial f /\partial x_i}, \forall j\neq i,
$$ 
assuming the RHS denominator is not zero. Now, assuming that the minimum of $x_i$ is obtained in the interior of the domain of $\{x_j\}_{j\neq i}$, i.e., the minimum of $x_i$ is achieved on a surface point where the other coordinates do not have their minimum or maximum value, one may find the minimum of $x_i$ by examining its stationary points obtained by solving the system of equations 
$$
\left\{\begin{align}
\frac{\partial x_i}{\partial x_j}&= 0,\forall j\neq i,\\
f&=0
\end{align}\right\} .
$$
Again, considering $f = \sum_{i=1}^nx_i^2-c$ as an example, it follows that 
$$
\frac{\partial x_i}{\partial x_j}=-\frac{x_j}{x_i},
$$
whenever $x_i\neq 0$. Assuming the latter holds, it directly follows that the above derivative is zero if and only if $x_j=0$, which in turn leads to identifying the stationary points for $x_i$ as $+\sqrt{c}$ and $-\sqrt{c}$. Clearly, the minimum corresponds to $-\sqrt{c}$ (which is also smaller than the problematic value $x_i=0$, hence no need to further investigate the potential of $x_i=0$ as the minimum value).
