Lagrange multipliers

To apply it, set up the system \(f_x=\lambda g_x\), \(f_y=\lambda g_y\), \(g(x,y)=k\): three equations in the unknowns \(x\), \(y\), \(\lambda\). Solve it, evaluate \(f\) at each solution and compare; the largest value is the maximum and the smallest the minimum. If the constraint is a closed, bounded curve, both exist.

Guiding example: maximize \(xy\) with \(x+y=10\). The equations \(y=\lambda\), \(x=\lambda\) give \(x=y\), and with the constraint \(x=y=5\), \(f=25\). The method does not by itself tell whether a solution is a maximum or a minimum: you must compare the values of \(f\).

The geometric idea is that on the curve \(g=k\) an extremum occurs where a level curve of \(f\) touches it without crossing; there both are tangent and, since each gradient is perpendicular to its level curve, \(\nabla f\) and \(\nabla g\) are parallel. If they were not, you could move along the curve and increase or decrease \(f\).

The method requires \(\nabla g\ne\mathbf 0\) at the points of the constraint; points where \(\nabla g=\mathbf 0\) must be checked separately. With two constraints you use two multipliers, \(\nabla f=\lambda\nabla g+\mu\nabla h\). The value of \(\lambda\) measures how much the optimum changes if \(k\) is relaxed slightly.