Lagrange Multipliers
The Lagrange multiplier theorem roughly states that at any stationary point of the function that also satisfies the equality constraints, the gradient of the function at that point can be expressed as a linear combination of the gradients of the constraints at that point, with the Lagrange multipliers acting as coefficients. The relationship between the gradient of the function and gradients of the constraints leads to Lagrangian function.
Example:
Minimize a function of two variables z = xy under a constraint qiven by equality c: .
The great advantage of this method is that it allows the optimization to be solved without explicit parameterization in terms of the constraints.
Let us denote function z = f(x, y) and the constraints given by equality as g(x, y) = 0.
Lagrangian function L(x, y) = f(x, y) - λ g(x, y)
Stationary point of the Lagrangian function: L'x (x, y) = 0 andL'y (x, y) = 0 gives a necessary condition for optimality in constrained problems.
For the hyperboloid z = x . y above the hyperbola x2
− 3xy + y2= 20 we have Langrangian function
L(x, y) = x . y - λ (x2 − 3xy + y2 − 20)
Equations for stationary point
L'x (x, y) = y − 2λx + 3λy = 0 andL'y (x, y) = x + 3λx − 2λy = 0
Add equations together and factor the member (x+y). From the acquired reduction we have the relation for minimal points: x = − y.
Conclusion
Function z = x.y has two minimal values above hyperbola: at points Min1 = (−2, 2) and Min2 = (2, −2). Task 2: Paraboloid constrained along the circle
Find the extrema points of a function z = xy subject to equality constraint .
Parameterization of the circle gives us the function z(t) of the height over the circle.
z = xy = cos t . sin t
Stationary points: z'=0 ⇒
Maxima:
Minima: