Context
SVM: From Primal to Dual Primal form: $$\min_{w, b} \frac{1}{2} \|w\|^2 + C \sum_{i=1}^M \xi_i$$ subject to $y^{(i)}(w^T x^{(i)} + b) \geq 1 - \xi_i$, $i = 1, \ldots, M$, and $\xi_i \geq 0$, $i = 1, \ldots, M$.
We form the Lagrangian: $$L(w, b, \xi, \alpha, \beta) = \frac{1}{2} \|w\|^2 + C \sum_{i=1}^M \xi_i - \sum_{i=1}^M \alpha_i[y^{(i)}(w^T x^{(i)} + b) - 1 + \xi_i] - \sum_{i=1}^M \beta_i \xi_i$$
SVM: From Primal to Dual Take the derivatives of $L$ with respect to $w$, $b$, and $\xi$ and set them to zero: $$\nabla_w L = w - \sum_{i=1}^M \alpha_i y^{(i)} x^{(i)} = 0$$ $$\nabla_b L = \sum_{i=1}^M \alpha_i y^{(i)} = 0$$ $$\nabla_{\xi_i} L = C - \alpha_i - \beta_i = 0, \quad i = 1, \ldots, M$$
Plug them back into the Lagrangian: $$g(\alpha) = \sum_{i=1}^M \alpha_i - \frac{1}{2} \sum_{i,j=1}^M y^{(i)} y^{(j)} \alpha_i \alpha_j (x^{(i)})^T x^{(j)}$$
SVM: The Dual Problem Put this together with the constraints and eliminate $\beta_i$: $$\max_\alpha \sum_{i=1}^M \alpha_i - \frac{1}{2} \sum_{i,j=1}^M y^{(i)} y^{(j)} \alpha_i \alpha_j (x^{(i)})^T x^{(j)}$$ subject to $ \sum_{i=1}^{M} \alpha_i y^{(i)} = 0, $ $ 0 \leq \alpha_i \leq C, \quad i = 1, \ldots, M $
Problem
The original primal problem is minimization problem. According to the equation of Lagrangian function, with higher Lagrangian values, It should mean that there are more constraints violated, Why should we take $\max_\alpha$ at the last line instead of $\min_\alpha$ ?