However, we can add to $$ df = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz, $$ a multiplum of \( d\phi \), viz. \( \lambda d\phi \), resulting in $$ df+\lambda d\phi = (\frac{\partial f}{\partial z}+\lambda \frac{\partial \phi}{\partial x})dx+(\frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y})dy+ (\frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z})dz =0. $$ Our multiplier is chosen so that $$ \frac{\partial f}{\partial z}+\lambda\frac{\partial \phi}{\partial z} =0. $$
We need to remember that we took \( dx \) and \( dy \) to be arbitrary and thus we must have $$ \frac{\partial f}{\partial x}+\lambda\frac{\partial \phi}{\partial x} =0, $$ and $$ \frac{\partial f}{\partial y}+\lambda\frac{\partial \phi}{\partial y} =0. $$ When all these equations are satisfied, \( df=0 \). We have four unknowns, \( x,y,z \) and \( \lambda \). Actually we want only \( x,y,z \), \( \lambda \) needs not to be determined, it is therefore often called Lagrange's undetermined multiplier. If we have a set of constraints \( \phi_k \) we have the equations $$ \frac{\partial f}{\partial x_i}+\sum_k\lambda_k\frac{\partial \phi_k}{\partial x_i} =0. $$