Given the data points $x_1, \ldots, x_n \in \mathbb{R}^d$ and labels $y_1, \ldots, y_n \in \left \{-1, 1 \right\}$, the hard margin SVM primal problem is

$$ \text{minimize}_{w, w_0} \quad \frac{1}{2} w^T w $$ $$ \text{s.t.} \quad \forall i: y_i (w^T x_i + w_0) \ge 1$$

which is a quadratic program with $d+1$ variables to be optimized for and $i$ constraints. The dual

$$ \text{maximize}_{\alpha} \quad \sum_{i=1}^{n}{\alpha_i} - \frac{1}{2}\sum_{i=1}^{n}{\sum_{j=1}^{n}{y_i y_j \alpha_i \alpha_j x_i^T x_j}}$$ $$ \text{s.t.} \quad \forall i: \alpha_i \ge 0 \land \sum_{i=1}^{n}{y_i \alpha_i} = 0$$ is a quadratic program with $n + 1$ variables to be optimized for and $n$ inequality and $n$ equality constraints.

When implementing a hard margin SVM, why would I solve the dual problem instead of the primal problem? The primal problem looks more 'intuitive' to me, and I don't need to concern myself with the duality gap, the Kuhn-Tucker condition etc.

It would make sense to me to solve the dual problem if $d \gg n$, but I suspect there are better reasons. Is this the case?