Question

If, instead of \(g(x, y)=c,\) the constraint is written in the form of \(C(x, y)=0,\) how should the Lagrangian function and the first-order condition be modified as a consequence?

Short answer

The Lagrangian becomes \(\mathcal{L}(x, y, \lambda) = f(x, y) + \lambda C(x, y)\). First-order conditions are \(\frac{\partial \mathcal{L}}{\partial x} = 0\), \(\frac{\partial \mathcal{L}}{\partial y} = 0\), and \(C(x, y) = 0\).

Step-by-step solution

Identifying the New Constraint Form

In the problem, we change the constraint from being equal to a constant, \(g(x, y) = c\), to being equal to zero, \(C(x, y) = 0\). This alters how we construct the Lagrangian.

Constructing the Lagrangian

For the given problem, the Lagrangian, \(\mathcal{L}\), is modified to include the zero-form constraint as \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda C(x, y) \), where \(\lambda\) is the Lagrange multiplier.

First-Order Conditions with Modified Constraint

To derive the first-order conditions, we find the partial derivatives of the Lagrangian concerning \(x\), \(y\), and \(\lambda\). Solve the equations: \( \frac{\partial \mathcal{L}}{\partial x} = \frac{\partial f}{\partial x} + \lambda \frac{\partial C}{\partial x} = 0\), \(\frac{\partial \mathcal{L}}{\partial y} = \frac{\partial f}{\partial y} + \lambda \frac{\partial C}{\partial y} = 0 \), along with the constraint \(C(x, y) = 0\).

Revising the Approach

Note how this revision essentially shifts interpreting the constraint implicitly in modifying each part of the Lagrangian and affects solving the derivatives, accommodating a scenario where our constraint is no longer a constant equals expression.

Key concepts

Constraint Transformation

When working with optimization problems, constraints can sometimes be represented in different forms. Originally, a constraint might be given as an equation that equals a constant, for example, \(g(x, y) = c\). However, in some cases, we see it represented as \(C(x, y) = 0\). This transformation is a common technique in Lagrangian optimization.

• Changing the constraint from a constant to zero doesn't alter the problem's essence but helps simplify certain mathematical operations.
• The zero-form constraint \(C(x, y) = 0\) becomes particularly useful in analysis, as it means we do not need an extra equation to equate terms to a constant.
• This form is more versatile and aids in systematically constructing the Lagrangian function, as seen in mathematical derivations.

By understanding how to transform constraints, you can interpret and manipulate problems more effectively, especially when different forms can provide computational advantages.

First-Order Conditions

The first-order conditions are critical in finding the optimal solutions of a problem, especially when using the Lagrangian method. To find them, we derive the Lagrangian's partial derivatives with respect to each variable involved, including the Lagrange multiplier.

• The process involves forming equations by setting these partial derivatives equal to zero, indicating potential optimal points.
• For instance, with \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda C(x, y) \), you calculate \( \frac{\partial \mathcal{L}}{\partial x} \), \( \frac{\partial \mathcal{L}}{\partial y} \), and potentially \( \frac{\partial \mathcal{L}}{\partial \lambda} \).
• The result is a system of equations that includes the derivatives and the transformed constraint equation, \(C(x, y) = 0\).

First-order conditions are essential for pinpointing where the function reaches extreme values, thereby focusing efforts on areas that might provide optimal solutions.

Lagrange Multipliers

Lagrange multipliers are a central feature in constrained optimization problems. They provide a method to incorporate constraints into the optimization objective.

• In problems where constraints are present, the Lagrange multiplier \(\lambda\) acts like a weighting factor, allowing constraints to influence the function being optimized.
• Including \(\lambda\) helps ensure any changes in the function's desired outcome respect the constraint \(C(x, y) = 0\).
• Essentially, they reflect how much the objective function could be improved by "relaxing" the constraint slightly, offering insight into the constraint's trade-off.

Overall, Lagrange multipliers bridge the gap between pure calculation and real-world constraints, ensuring that optimal solutions aren't just perfect in theory but practical as well.