Fal!2001 16.31 Optimization -2 To find the constrained minimum form the augmented cost function L,f(x,y)+λc(x,y)=x2+y2+入(x+y+2) Where X is the lagrange multiplier e Note that if the constraint is satisfied, then L= f The solution approach without constraints is to find the stationary point of f(a, y)(af ax=af ay=0) With constraints we find the stationary points of L OL alaL axya入 which gives OL 2x+入=0 a=20+X=0 OL +y+2=0 e This gives 3 equations in 3 unknowns, solve to fine The key point here is that due to the constraint, the selection of a and y during the minimization are not independent The Lagrange multiplier captures this dependency The lQr optimization follows the same path as this, but it is com- plicated by the fact that the cost involves an integration over timeFall 2001 16.31 Optimization-2 • To find the constrained minimum, form the augmented cost function L , f(x, y) + λc(x, y) = x2 + y2 + λ(x + y + 2) — Where λ is the Lagrange multiplier • Note that if the constraint is satisfied, then L ≡ f • The solution approach without constraints is to find the stationary point of f(x, y) (∂f/∂x = ∂f/∂y = 0) — With constraints we find the stationary points of L ∂L ∂L ∂L = = = 0 ∂x ∂y ∂λ which gives ∂L ∂x = 2x + λ = 0 ∂L = 2y + λ = 0 ∂y ∂L = x + y + 2 = 0 ∂λ • This gives 3 equations in 3 unknowns, solve to find x? = y? = −1 • The key point here is that due to the constraint, the selection of x and y during the minimization are not independent — The Lagrange multiplier captures this dependency. • The LQR optimization follows the same path as this, but it is com￾plicated by the fact that the cost involves an integration over time