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.