Problem 11 - Calculus an Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \begin{array}{ll}{\text { Objective Function }} & {\text { Constraint }} \\\ {\text { Maximize } f(x, y)=e^{x y}} & {x^{2}+y^{2}-8=0}\end{array} $$

Step-by-Step Solution

Answer

The solution to this exercise requires solving a system of three equations, which involves certain mathematical manipulation. The final solution will be a set of values for \(x\), \(y\), and λ.

Step 1: Set up the Lagrange system

Firstly, set up the Lagrange system. The equation for the Lagrange system is: \(f(x, y) - λg(x, y) = 0\), where \(f(x, y)\) is the objective function, \(g(x, y)\) is the constraint, and λ is the Lagrange multiplier. Substituting the given functions gives the following system of equations: \[e^{xy} - λ(2x) = 0\] \[xe^{xy} - λ(2y) = 0\] \(x^{2}+y^{2}-8 = 0.\]

Step 2: Solve the Lagrange systeme

Next, solve the system of equations. As these equations can be quite complex, they can potentially require some algebraic manipulation to simplify. Also, there will probably be more than one solution, as these are simultaneous nonlinear equations. Simplify the first two equations and solve for λ: \[λ = \frac{e^{xy}}{2x} = \frac{xe^{xy}}{2y}.\] Solving for x or y will yield the ratio between x and y. Substitute to the third equation to find actual values.

Step 3: Verify the solution

Once the potential points are located, they should be verified if they indeed give a maximum, minimum or saddle point. Evaluate the objective function at the solutions derived to determine which value is maximum.

Key Concepts

Objective Function

The objective function is a central element in optimization problems. In our exercise, this is the expression we want to maximize or minimize. Here, the objective function is given by \( f(x, y) = e^{xy} \), which we aim to maximize.
This function describes a surface in three-dimensional space, where the height represents the value of \( f \). Our goal is to find the point on this surface that gives the highest value, constrained by another condition.

Remember, an objective function can take many forms, and it often represents something you want to optimize, like profit or efficiency. In mathematics, it's tied to the decision variables—in this case, \( x \) and \( y \)—which you have control over, within specified constraints.

Constraint

Constraints are rules or limits placed on the variables in an optimization problem. They are essential because they define the feasible region for solutions. In this exercise, the constraint is \( x^2 + y^2 - 8 = 0 \), meaning our solutions must lie on the circle defined by this equation.
Constraints can significantly affect the outcome of an optimization problem. They ensure that solutions remain within realistic or desired boundaries, reflecting real-world limits like financial budgets or physical limitations.
When working with Lagrange multipliers, each constraint comes with its multiplier, λ, that accounts for how sensitive the optimal solution is to changes in that constraint. Constraints can be equalities, like in this exercise or inequalities, narrowing down the possible solutions.

Extremum

An extremum refers to the maximum or minimum value of a function within a defined region. In optimization, finding an extremum is the main goal. In the context of our exercise, we specifically aim to maximize the function \( f(x, y) = e^{xy} \).
Extremum can be local or global. A local extremum is the highest or lowest point in a small region, while a global extremum is the highest or lowest point in the entire domain. Identifying whether a point is a maximum, minimum, or saddle point often requires analyzing the nature of the function around that point.
In the Lagrange multipliers method, after solving the system of equations, the potential extremum points need to be verified to ensure they provide maximum or minimum values, as required by the problem.

Nonlinear Equations

Nonlinear equations are those that do not form a straight line when graphed or cannot be expressed by a linear equation of the form \( ax + by + c = 0 \). In this exercise, the equations derived from the Lagrange multipliers method form a nonlinear system:

• \( e^{xy} - λ(2x) = 0 \)
• \( xe^{xy} - λ(2y) = 0 \)
• \( x^2 + y^2 - 8 = 0 \)

Solving such equations can be complex and may require iterative or numerical methods for solutions. These equations must be manipulated algebraically to isolate variables or parameters, like λ in this context, to progress toward a practical solution.
Nonlinear systems can have multiple or even no solutions, making them challenging but interesting. They are notable for their ability to model real-world problems more accurately than linear systems, which only provide approximations.