Example how maximizing and minimizing a function can be equivalent? I don't understand how sometimes given an optimization problem, a function could get its optimal solution by minimizing or sometimes just by reformulation it becomes maximizing. Can you please give me an example for this?
 A: Consider a function $f$ which for the purpose of this discussion, I am going to take to be positive. The aim with optimization is to find the values of the argument(s) that make the function achieve its 'best' value (if you're maximizing, to find the argmax).
Here's an example showing an $f$ we want to maximize, along with $\log f$, $-f$ and $1/f$. Now, any time you move an $x$ with some value of $f$ to another $x$ with a higher value of $f$, any strictly monotonic increasing transformation of $f$ (e.g. $\log$) will also increase, and any monotonic decreasing transformation of $f$ will decrease (e.g. $-f$, $1/f$). That is, if $f(x_2)>f(x_1)$, then $\log f(x_2)> \log f(x_1)$, and $-f(x_2)< -f(x_1)$, and so on (as long as they're all finite). 
As long as all the transformed values must also be finite, if $f$ achieves some optimum at $x^*$, any monotonic increasing function will, too. So $f$ and $\log f$ will have maxima at the same $x$-value, and $-f$ and $1/f$ with have minima there:

The value of $x$ that causes $f$ to be at its maximum is marked in ($x=5$), and the corresponding position is marked for $\log f$, $-f$ and $1/f$; the same value of $x$ ($x=5$) is the argmax or argmin as appropriate.
(In this example, $f$ happened to be a logistic density, with mean $5$.)
A: Sometimes the optimization function is converted several times for simplicity. 
Take SVM for example. The original objective function is:

It is just equivalent to:

And this can be solved with Quadratic Programming package. 
With Lagrange duality, the objective function can be further converted to:

Let , The objective function is now changed to:
     (1)
Now consider another objective that just changes the order of maximum and minimum (also called the dual of (1)):
     (2)
Note that  ≤ . When Karush-Kuhn-Tucker condition is satisfied, the solution of (2) is equal to the one of (1), and (2) is easier to solve since after the derivative on $w$ and $b$ to minimize the function, there is only one unknown $\alpha$ in the optimization function.
From the example above you can notice there are several times in one problem that the minimization and maximization can be exchanged in order to get the solution more quickly. 
