Why do we try to maximize Lagrangian in SVMs?
I think the answer to this deals with convex functions and duality. 

$$L(w, b, \alpha) = \frac{1}{2}\|w\|^2 + \sum_i α_i(y_i(w \bullet x+b)-1).$$ 

When you minimize this, you are minimizing it over $w,b$. As you stated, this will produce the Lagrangian dual function:

$\Theta (\alpha) = inf L(w, b, \alpha) = \sum_iα_i-\frac{1}{2}\sum_{i,j}y_iy_jα_iα_j(x_i\bullet x_j)$

The maximization must be done here, but of the function $\Theta (\alpha)$ (the Lagrangian dual function).

Here is some background on why we are maximizing:

1) Let $p*$ be the optimal value of the problem of minimizing $\frac{1}{2}\|w\|^2$ (the primal). The Lagrangian dual function has the property that $L(w, b, \alpha) \leq p*$. It is a lower bound on the primal function. Instead of solving the primal problem, we are want to get the maximum lower bound on $p*$ by maximizing the Lagrangian dual function (the dual problem).

2) The primal problem is a convex function. The Lagrangian dual function is a concave function(can be verified). Maximizing a concave function is equivalent to minimizing a convex function. Therefore it is easy.

3) Why maximizing $\Theta (\alpha)$ is as good as minimizing $\frac{1}{2}\|w\|^2$: Let $d*$ be the value of $max \Theta (\alpha)$. This primal convex optimization problem satisfies Slater's constraint qualification which gives us strong duality ($d*=p*$). Therefore, $max \Theta (\alpha)$ is the same as $min \frac{1}{2}\|w\|^2$.