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where <math>(\lambda_k)_k</math> and <math>(\mu_j)_j</math> are the vectors of multipliers. Taking the partial derivative of the Lagrangian with respect to each good <math>x_j^k</math> for <math>j=1,\ldots,n</math> and <math>k=1,\ldots, m</math> and gives the following system of first-order conditions:

where <math>(\lambda_k)_k</math> and <math>(\mu_j)_j</math> are the vectors of multipliers. Taking the partial derivative of the Lagrangian with respect to each good <math>x_j^k</math> for <math>j=1,\ldots,n</math> and <math>k=1,\ldots, m</math> and gives the following system of first-order conditions:

其中<math>(\lambda_k)_k</math>和<math>(\mu_j)_j</math>是乘子的向量。取关于商品的拉格朗日函数的偏导数<math>x_j^k</math>，其中<math>j=1,\ldots,n</math> ，<math>k=1,\ldots, m</math>，并给出以下一阶条件系统:

其中<math>(\lambda_k)_k</math>和<math>(\mu_j)_j</math>是乘子的向量。取关于商品的拉格朗日函数的偏导数<math>x_j^k</math>（<math>j=1,\ldots,n</math> ，<math>k=1,\ldots, m</math>）并给出以下一阶条件系统:

:对于<math>j=1,\ldots,n</math>

对于<math>j=1,\ldots,n</math>

::<math>\frac{\partial L_i}{\partial x_j^i} = f_{x^i_j}^1-\mu_j=0\</math>

:<math>\frac{\partial L_i}{\partial x_j^i} = f_{x^i_j}^1-\mu_j=0\</math>

:对于<math>k= 2,\ldots,m</math>，<math>j=1,\ldots,n</math>

对于<math>k= 2,\ldots,m</math> ， <math>j=1,\ldots,n</math>

::<math>\frac{\partial L_i}{\partial x_j^k} = -\lambda_k f_{x^k_j}^i-\mu_j=0</math>

:<math>\frac{\partial L_i}{\partial x_j^k} = -\lambda_k f_{x^k_j}^i-\mu_j=0</math>