Steps to Apply the Method

1. Formulate the Lagrangian Function

The Lagrangian function \( \mathcal{L} \) is defined as:

\[ \mathcal{L}(x, y, z, \lambda) = f(x, y, z, \dots) - \lambda g(x, y, z, \dots) \] where:

\( f(x, y, z, \dots) \) is the function to optimize,

\( g(x, y, z, \dots) = 0 \) is the constraint,

\( \lambda \) is the Lagrange multiplier.

2. Solve the System of Equations

Find the points where the gradient of \( \mathcal{L} \) is zero by solving:

\[ \nabla \mathcal{L} = 0 \] This expands into: \[ \nabla f(x, y, z, \dots) = \lambda \nabla g(x, y, z, \dots) \] and \[ g(x, y, z, \dots) = 0. \]

3. Identify the Critical Points

The solutions to the above system are the critical points. Evaluate \( f(x, y, z, \dots) \) at these points.

4. Compare Values

Compare the values of \( f(x, y, z, \dots) \) at all critical points to determine the maximum and/or minimum values.

Example

Let \( f(x, y) = x^2 + y^2 \) be the function to maximize/minimize, subject to the constraint \( g(x, y) = x + y - 1 = 0 \).

Step 1: Write the Lagrangian

\[ \mathcal{L}(x, y, \lambda) = x^2 + y^2 - \lambda (x + y - 1) \]

Step 2: Compute the Gradient

Find \( \nabla \mathcal{L} \) and set it to zero:

\[ \nabla \mathcal{L} = \left( \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial \lambda} \right) = 0 \] This gives: \[ \frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda = 0 \quad \Rightarrow \quad \lambda = 2x \] \[ \frac{\partial \mathcal{L}}{\partial y} = 2y - \lambda = 0 \quad \Rightarrow \quad \lambda = 2y \] \[ \frac{\partial \mathcal{L}}{\partial \lambda} = -(x + y - 1) = 0 \quad \Rightarrow \quad x + y = 1 \]

Step 3: Solve the System

From \( \lambda = 2x \) and \( \lambda = 2y \), we get \( 2x = 2y \), or \( x = y \). Substituting \( x = y \) into \( x + y = 1 \), we find \( x = y = \frac{1}{2} \).

Step 4: Evaluate \( f(x, y) \)

At \( (x, y) = \left(\frac{1}{2}, \frac{1}{2}\right) \): \[ f(x, y) = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}. \] Thus, the maximum value of \( f(x, y) \) subject to the constraint is \( \frac{1}{2} \).