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The Lagrangian function is a powerful tool used to find the extrema of a function subject to constraints—essentially, it helps us optimize a given function under specific conditions. When dealing with an optimization problem, you might encounter an objective function, such as \( f(x, y) = x + 2y \), which you wish to maximize or minimize. However, this optimization is subject to a constraint, expressed by another equation, for instance, \( x^2 + y^2 = 4 \).

In these scenarios, we introduce an auxiliary variable, called a Lagrange multiplier (denoted by \( \lambda \)), and construct the Lagrangian function, which incorporates both our objective function and the constraint. The general form of the Lagrangian function is \( L(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - c) \), where \( g(x, y) \) represents the constraint, and \( c \) is a constant. By setting up the Lagrangian function properly, we establish a foundation for using partial derivatives to solve for possible extrema.

To find the stationary points of the Lagrangian function, we need to understand and compute partial derivatives. Partial derivatives represent the rate of change of a multivariable function with respect to one variable while holding others constant. For example, in the context of the Lagrangian function \( L(x, y, \lambda) \), we calculate the partial derivatives with respect to \( x \), \( y \), and \( \lambda \).

Computing \( \frac{\partial L}{\partial x} \), \( \frac{\partial L}{\partial y} \), and \( \frac{\partial L}{\partial \lambda} \) allows us to set up a system of equations—each equal to zero which embodies the critical points where the rate of change is flat. These derivatives lay out the pathway to solve for the values of \( x \), \( y \), and \( \lambda \) that satisfy the optimization problem.

Stationary points, or critical points, are the solutions that we find when the partial derivatives of the Lagrangian function equal zero. These points are where the function neither increases nor decreases, suggesting possible locations of maximums, minimums, or saddle points. After solving the system of equations derived from setting the partial derivatives to zero, we obtain a set of potential solutions.

In our example, setting the equations \( 1 - 2\lambda x = 0 \), \( 2 - 2\lambda y = 0 \), and \( x^2 + y^2 - 4 = 0 \) leads to four stationary points: \( (x, y) = (\pm\sqrt{2}, \pm\sqrt{2}) \). These points serve as candidates that require further examination to determine if they represent maximum or minimum values of the original function under the given constraint.

Once we've identified the stationary points, the final step is to evaluate the original function at each point to determine which corresponds to the maximum and minimum values. This step helps us conclude which set of variables provides the highest or lowest outcome for the objective function when the constraint is factored in.

By substituting the stationary points into \( f(x, y) \), we can extract the function values: \( f(\sqrt{2}, \sqrt{2}) = 3\sqrt{2} \), and \( f(-\sqrt{2}, -\sqrt{2}) = -3\sqrt{2} \), revealing that the highest and lowest function values are \( 3\sqrt{2} \) and \( -3\sqrt{2} \), respectively. These evaluations conclude the optimization problem: the maximum value of \( f(x, y) \) subject to the constraint \( x^2 + y^2 = 4 \) is \( 3\sqrt{2} \), while the minimum value is \( -3\sqrt{2} \).