The margin between the separating hyperplane and the class boundaries of an SVM is an essential feature of this algorithm.

See, you have two hyperplanes (1) w^tx+b>=1, if y=1 and (2) w^tx+b<=-1, if y=-1. This says that any vector with a label y=1 must lie ether on or behind the hyperplane (1). The same applies to the vectors with label y=-1 and hyperplane (2).

Note: If those requirements can be fulfilled, it implicitly means the dataset is linearly separatable. This makes sense because otherwise no such margin can be constructed.

So, what an SVM tries to find is a decision boundary which ist half-way between (1) and (2). Let's define this boundary as (3) w^tx+b=0. What you see here is that (1), (2) and (3) are parallel hyperplanes because they share the same parameters w and b. The parameters w holds the direction of those planes. Recall that a vector always has a direction and a magnitude/length.

The question is now: How can one calculate the hyperplane (3)? The equations (1) and (2) tell us that any vector with a label y=1 which is closest to (3) lies exactly on the hyperplane (1), hence (1) becomes w^tx+b=1 for such x. The similar applies for the closest vectors with a negative label and (2). Those vectors on the planes called 'support vectors' and the decision boundary (3) only depends on those, because one simply can subtract (2) from (1) for the support vectors and gets:

w^tx+b-w^tx+b=1-(-1) => wt^x-w^tx=2

Note: x for the two planes are different support vectors.

Now, we want to get the direction of w but ignoring it's length to get the shortest distance between (3) and the other planes. This distance is a perpendicular line segment from (3) to the others. To do so, one can divide by the length of w to get the norm vector which is perpendicular to (3), hence (wt^x-w^tx)/||w||=2/||w||. By ignoring the left hand site (it's equal) we see that the distance between the two planes is actually 2/||w||. This distance must be maximized.

Edit: As others state here, use Lagrange multipliers or the SMO algorithm to minimize the term 1/2 ||w||^2 s.t. y(w^tx+b)>=1 This is the convex form of the optimization problem for the primal svm.

Let $\textbf{x}_+$ be a positive example on one gutter, such that $$\textbf{w} \cdot \textbf{x}_+ - b = 1$$

Let $\textbf{x}_-$ be a negative example on another gutter, such that $$\textbf{w} \cdot \textbf{x}_- - b = -1$$

The width of margin is the scalar projection of $\textbf{x}_+ - \textbf{x}_-$ on unit normal vector , that is the dot production of $\textbf{x}_+ - \textbf{x}_-$ and $\frac{\textbf{w}}{\|\textbf{w}\|}$

\begin{align} width & = (\textbf{x}_+ - \textbf{x}_-) \cdot \frac{\textbf{w}}{\|\textbf{w}\|} \\ & = \frac {(\textbf{x}_+ - \textbf{x}_-) \cdot {\textbf{w}}}{\|\textbf{w}\|} \\ & = \frac{\textbf{x}_+ \cdot \textbf{w} \,{\bf -}\, \textbf{x}_-\cdot \textbf{w}}{\|\textbf{w}\|} \\ & = \frac{1-b-(-1-b)}{\lVert \textbf{w} \rVert} \\ & = \frac{2}{\|\textbf{w}\|} \end{align}

The above refers to MIT 6.034 Artificial Intelligence

The functional margin represents the correctness and confidence of the prediction if the magnitude of the vector(w^T) orthogonal to the hyperplane has a constant value all the time.

By correctness, the functional margin should always be positive, since if wx + b is negative, then y is -1 and if wx + b is positive, y is 1. If the functional margin is negative then the sample should be divided into the wrong group.

By confidence, the functional margin can change due to two reasons: 1) the sample(y_i and x_i) changes or 2) the vector(w^T) orthogonal to the hyperplane is scaled (by scaling w and b). If the vector(w^T) orthogonal to the hyperplane remains the same all the time, no matter how large its magnitude is, we can determine how confident the point is grouped into the right side. The larger that functional margin, the more confident we can say the point is classified correctly.

But if the functional margin is defined without keeping the magnitude of the vector(w^T) orthogonal to the hyperplane the same, then we define the geometric margin as mentioned above. The functional margin is normalized by the magnitude of w to get the geometric margin of a training example. In this constraint, the value of the geometric margin results only from the samples and not from the scaling of the vector(w^T) orthogonal to the hyperplane.

The geometric margin is invariant to the rescaling of the parameter, which is the only difference between geometric margin and functional margin.

EDIT:

The introduction of functional margin plays two roles: 1) intuit the maximization of geometric margin and 2) transform the geometric margin maximization issue to the minimization of the magnitude of the vector orthogonal to the hyperplane.

Since scaling the parameters w and b can result in nothing meaningful and the parameters are scaled in the same way as the functional margin, then if we can arbitrarily make the ||w|| to be 1(results in maximizing the geometric margin) we can also rescale the parameters to make them subject to the functional margin being 1(then minimize ||w||).