8.2 The method of Lagrange multipliers

Lagrange multipliers are a powerful tool for solving constrained optimization problems. They allow us to find the maximum or minimum values of a function subject to specific constraints, turning complex problems into simpler ones.

This method introduces a new variable, λ, which helps balance the objective function and constraints. By using Lagrange multipliers, we can tackle real-world problems like maximizing profits or minimizing costs while working within given limitations.

Lagrange Multipliers and Lagrangian Function

Lagrange Multiplier and Lagrangian Function

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• Lagrange multiplier ($\lambda$) introduced to solve constrained optimization problems
• Converts a constrained optimization problem into an unconstrained one by incorporating the constraints into the objective function
• Lagrangian function ($L$) formed by combining the objective function $f(x, y)$ and the constraint function $g(x, y)$ using Lagrange multipliers
  • $L(x, y, \lambda) = f(x, y) + \lambda g(x, y)$
  • Example: Minimize $f(x, y) = x^2 + y^2$ subject to $g(x, y) = x + y - 1 = 0$
    • Lagrangian function: $L(x, y, \lambda) = x^2 + y^2 + \lambda(x + y - 1)$

Critical Points and Lagrange Multiplier Theorem

• Critical points of the Lagrangian function found by setting the partial derivatives equal to zero
  • $\frac{\partial L}{\partial x} = 0$, $\frac{\partial L}{\partial y} = 0$, and $\frac{\partial L}{\partial \lambda} = 0$
  • Solving this system of equations yields the critical points $(x, y, \lambda)$
• Lagrange multiplier theorem states that if $(x_0, y_0)$ is a local extremum of $f(x, y)$ subject to the constraint $g(x, y) = 0$, then there exists a Lagrange multiplier $\lambda_0$ such that $(x_0, y_0, \lambda_0)$ is a critical point of the Lagrangian function
  • Provides a necessary condition for constrained optimization problems
  • Example: For the problem above, the critical points are $(0, 1, -2)$ and $(1, 0, -2)$

Gradient Vectors and Parallel Gradients

Gradient Vectors and Parallel Gradients

• Gradient vector $\nabla f(x, y)$ is a vector perpendicular to the level curve of $f(x, y)$ at a given point
  • $\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$
• At the optimal solution, the gradient vectors of the objective function and the constraint function are parallel
  • $\nabla f(x_0, y_0) = \lambda \nabla g(x_0, y_0)$
  • Geometrically, this means that the level curves of $f(x, y)$ and $g(x, y)$ are tangent at the optimal point

First-Order Necessary and Second-Order Sufficient Conditions

• First-order necessary conditions for a point $(x_0, y_0)$ to be a local extremum of $f(x, y)$ subject to $g(x, y) = 0$
  1. $g(x_0, y_0) = 0$ (the point satisfies the constraint)
  2. $\nabla f(x_0, y_0) = \lambda \nabla g(x_0, y_0)$ for some $\lambda$ (gradient vectors are parallel)
• Second-order sufficient conditions help classify the critical points as local minima, local maxima, or saddle points
  • Involves examining the Hessian matrix of the Lagrangian function at the critical points
  • Example: For the problem above, $(0, 1, -2)$ is a local minimum, and $(1, 0, -2)$ is a local maximum

Constrained Optimization

Saddle Points and Multiple Constraints

• Saddle points are critical points that are neither local minima nor local maxima
  • Occur when the Hessian matrix of the Lagrangian function has both positive and negative eigenvalues
  • Represent points where the objective function increases in some directions and decreases in others
• Problems with multiple constraints can be solved using Lagrange multipliers by introducing a Lagrange multiplier for each constraint
  • Lagrangian function: $L(x, y, \lambda_1, \lambda_2, \ldots, \lambda_n) = f(x, y) + \lambda_1 g_1(x, y) + \lambda_2 g_2(x, y) + \ldots + \lambda_n g_n(x, y)$
  • First-order necessary conditions: $\nabla f(x_0, y_0) = \lambda_1 \nabla g_1(x_0, y_0) + \lambda_2 \nabla g_2(x_0, y_0) + \ldots + \lambda_n \nabla g_n(x_0, y_0)$
  • Example: Minimize $f(x, y) = x^2 + y^2$ subject to $g_1(x, y) = x + y - 1 = 0$ and $g_2(x, y) = x - y - 1 = 0$
    • Lagrangian function: $L(x, y, \lambda_1, \lambda_2) = x^2 + y^2 + \lambda_1(x + y - 1) + \lambda_2(x - y - 1)$