Question

Write the Lagrangian function and the first-order condition for stationary values (without solving the equations) for each of the following: (a) \(z=x+2 y+3 w+x y-y w,\) subject to \(x+y+2 w=10\) (b) \(z=x^{2}+2 x y+y w^{2},\) subject to \(2 x+y+w^{2}=24\) and \(x+w=8\)

Short answer

The Lagrangian functions and first-order conditions are set for both parts, indicating stationary points using Lagrange multipliers.

Step-by-step solution

Understanding the Objective Function (a)

For part (a), we need to find the Lagrangian function of the objective function given by:\[ z = x + 2y + 3w + xy - yw \]This function involves the variables \(x\), \(y\), and \(w\).

Identifying the Constraint (a)

The constraint provided for part (a) is:\[ x + y + 2w = 10 \]

Writing the Lagrangian Function (a)

The Lagrangian function combines the objective function and the constraints using a Lagrange multiplier (\(\lambda\)):\[ \mathcal{L}(x, y, w, \lambda) = x + 2y + 3w + xy - yw - \lambda(x + y + 2w - 10) \]

First-Order Conditions (a)

To find the stationary values, set the partial derivatives of the Lagrangian with respect to each variable and the Lagrange multiplier to zero:1. \( \frac{\partial \mathcal{L}}{\partial x} = 1 + y - \lambda = 0 \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 2 + x - w - \lambda = 0 \)3. \( \frac{\partial \mathcal{L}}{\partial w} = 3 - y - 2\lambda = 0 \)4. \( \frac{\partial \mathcal{L}}{\partial \lambda} = -(x + y + 2w - 10) = 0 \)

Understanding the Objective Function (b)

For part (b), the objective function is:\[ z = x^2 + 2xy + yw^2 \]This function involves the variables \(x\), \(y\), and \(w\).

Identifying the Constraints (b)

There are two constraints for part (b):1. \( 2x + y + w^2 = 24 \)2. \( x + w = 8 \)

Writing the Lagrangian Function (b)

The Lagrangian function includes two Lagrange multipliers (\(\lambda_1\) and \(\lambda_2\)) for the two constraints:\[ \mathcal{L}(x, y, w, \lambda_1, \lambda_2) = x^2 + 2xy + yw^2 - \lambda_1(2x + y + w^2 - 24) - \lambda_2(x + w - 8) \]

First-Order Conditions (b)

Set the partial derivatives of the Lagrangian with respect to each variable and the Lagrange multipliers to zero:1. \( \frac{\partial \mathcal{L}}{\partial x} = 2x + 2y - 2\lambda_1 - \lambda_2 = 0 \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 2x - \lambda_1 = 0 \)3. \( \frac{\partial \mathcal{L}}{\partial w} = 2yw - 2w\lambda_1 - \lambda_2 = 0 \)4. \( \frac{\partial \mathcal{L}}{\partial \lambda_1} = -(2x + y + w^2 - 24) = 0 \)5. \( \frac{\partial \mathcal{L}}{\partial \lambda_2} = -(x + w - 8) = 0 \)

Summary of Conditions

For both parts (a) and (b), we have defined the Lagrangian functions and set the first-order conditions by equating the derivatives to zero. Solving these equations would give the stationary points, but solving is not required in this exercise.

Key concepts

Constrained Optimization

Constrained optimization is a mathematical approach used to find the maximum or minimum value of a function within a given set of restrictions or constraints. This method is crucial when variables have limitations or when operating under specific conditions.
In this exercise, we deal with functions involving variables and additional constraints that must hold true. For part (a), the constraint is that the sum of variables plus weighted components equals 10, i.e., \( x + y + 2w = 10 \). For part (b), we have two constraints: the first involves a combination of the variables and one squared term, while the second constraint directly links two variables: \( x + w = 8 \).
Using constrained optimization, we incorporate these restrictions into our problem-solving process. We use Lagrange multipliers, which help us manage and solve these problems by transforming them into a different form where both the objective function and the constraints are combined.

First-Order Conditions

First-order conditions are critical in determining where the objective function reaches its maximum or minimum value under the given constraints. These conditions rely on setting the partial derivatives of the Lagrangian function to zero.
For part (a), the Lagrangian function is derived from the objective and the constraint. By taking partial derivatives with respect to each variable \((x, y, w)\) and the Lagrange multiplier \(\lambda\), and setting them to zero, we form a system of equations. These equations help identify points that may be maxima or minima.
The same methodology applies for part (b), although with two constraints, two Lagrange multipliers are introduced. Again, taking partial derivatives and equating them to zero offers vital equations. Solving this system would lead to identifying the precise stationary points, which, however, isn't required for this specific task.

Objective Function

An objective function is the mathematical expression that we aim to maximize or minimize in the context of constrained optimization.
For part (a), our objective function is \( z = x + 2y + 3w + xy - yw \). We seek to find what combination of \(x, y, w\) will give us the highest or lowest possible value of \(z\), considering the constraint \(x + y + 2w = 10\).
In part (b), the objective function is different: \( z = x^2 + 2xy + yw^2 \). With two constraints influencing the outcome, our approach shifts to satisfying all conditions while optimizing this function. Understanding the objective function guides us in how to rewrite the problem into the Lagrangian form, which significantly influences the path to a solution.

Partial Derivatives

Partial derivatives play a vital role in Lagrange multipliers and constrained optimization by helping to locate points where the objective function has stationary values under given constraints.
When performing optimization, we take the partial derivative of the Lagrangian with respect to each variable and the Lagrange multipliers. For part (a), this involves derivatives \( \frac{\partial \mathcal{L}}{\partial x} \), \( \frac{\partial \mathcal{L}}{\partial y} \), and so on, all set to zero to find stationary positions.
Similarly, in part (b), because it involves multiple constraints, we take derivatives with respect to each variable \((x, y, w)\) and each Lagrange multiplier, yielding a system of equations. Solving these derivatives is crucial as they incorporate the effects of constraints into the process, ensuring coherence between the desired objective and the constraints.