This has in turn the consequences that we change our optmization problem to finding the minimum of $$ {\cal L}=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}-\sum_{i=1}^n\lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)-(1-\xi_)\right]+C\sum_{i=1}^n\xi_i-\sum_{i=1}^n\gamma_i\xi_i, $$ subject to $$ y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)=1-\xi_i \hspace{0.1cm}\forall i, $$ with the requirement \( \xi_i\geq 0 \).
Taking the derivatives with respect to \( b \) and \( \boldsymbol{w} \) we obtain $$ \frac{\partial {\cal L}}{\partial b} = -\sum_{i} \lambda_iy_i=0, $$ and $$ \frac{\partial {\cal L}}{\partial \boldsymbol{w}} = 0 = \boldsymbol{w}-\sum_{i} \lambda_iy_i\boldsymbol{x}_i, $$ and $$ \lambda_i = C-\gamma_i \hspace{0.1cm}\forall i. $$ Inserting these constraints into the equation for \( {\cal L} \) we obtain the same equation as before $$ {\cal L}=\sum_i\lambda_i-\frac{1}{2}\sum_{ij}^n\lambda_i\lambda_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j, $$ but now subject to the constraints \( \lambda_i\geq 0 \), \( \sum_i\lambda_iy_i=0 \) and \( 0\leq\lambda_i \leq C \). We must in addition satisfy the Karush-Kuhn-Tucker condition which now reads $$ \lambda_i\left[y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_)\right]=0 \hspace{0.1cm}\forall i, $$ $$ \gamma_i\xi_i = 0, $$ and $$ y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b) -(1-\xi_) \geq 0 \hspace{0.1cm}\forall i. $$