Finding minimum of multidimensional function My calculus knowledge is pretty limited, but unfortunately I need to solve a problem of the following kind:
I'm given a 2 dimensional function $f(x,y)$ from $\mathbb{R}^2$ to $\mathbb{R}$ and I want to know, where it attains its minimum value over $\mathbb{R}\times(a,b)$.
Put differently I want to find an $x$ value and a $y\in(a,b)$ such that $f(x,y) \leq f(x',y')$ for all x' in $\mathbb{R}$ and all $y \in (a,b)$.
I'll have to take the partial derivative of $f$ w.r.t $x$, but 
I don't understand how y will come into play.
 A: $f(x,y)$ has a critical point at $(x,y)$ if the gradient $\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$ is the zero vector at that point. 
So the procedure you'll want to follow is the following. I assume that $f$ does in fact attain its minium on $\mathbb{R} \times [a,b]$; if it's not obvious for your particular $f$, it's something you'll need to check.


*

*Find the points where the gradient of $f$ vanishes. Throw out critical points with $y$ not in $(a,b)$.

*Evaluate $f$ at these points to find the minimum on the interior of your region. (If there are no critical points in the region, skip this step.)

*Find the minimum of the one-dimensional functions $f(x,a)$ and $f(x,b)$. This will give you the minima at the boundary of your region.

*Evaluate $f$ at the three points from steps 2 and 3. Whichever gives the smallest $f$ is the global minimum.

A: If you are given an analytical function, then the partial derivative is a deterministic approach:
$\frac{\partial f}{\partial x}= 0$
$\frac{\partial f}{\partial y}= 0$
$\frac{\partial^2 f}{\partial x^2}> 0$
$\frac{\partial^2 f}{\partial y^2}> 0$
If you are given a "black box" function, then stochastic or Monte Carlo methods. If you want to go further, this book "Stochastic Adaptive Search for Global Optimization" (2003) is a good guide.
