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.