Problem 10 - Calculus an Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \begin{array}{ll}{\text { Objective Function }} & {\text { Constraint }} \\\ {\text { Maximize } f(x, y)=\sqrt{x^{2}+y^{2}}} & {2 x+4 y-15=0}\end{array} $$

Step-by-Step Solution

Answer

The maximum value of the given function is when \(x = \frac{15}{4}\) and \(y = \frac{15}{8}\) with a corresponding value of \(\frac{15\sqrt{2}}{4}\)

Step 1: Setting up the Lagrangian

The Lagrangian \(L\) can be written as \(L = f(x, y) - λ(g(x, y) - c)\), where \(f(x, y)\) is the function to be maximized, \(g(x, y)\) is the constraint function, \(c\) is a constant, and \(λ\) is the Lagrange multiplier. Here, \(f(x, y) = \sqrt{x^{2}+y^{2}}\), \(g(x, y) = 2x+4y\), and \(c = 15\). So, \(L = \sqrt{x^{2}+y^{2}} - λ(2x+4y-15)\).

Step 2: Computing the partial derivatives

Compute the partial derivatives of \(L\) with respect to \(x\), \(y\), and \(λ\) respectively and these are set to 0 to generate a system of equations. The partial derivatives are \(\frac{\partial L}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}}} - 2λ = 0\), \(\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^{2}+y^{2}}} - 4λ = 0\), and \(\frac{\partial L}{\partial λ} = 2x+4y-15 = 0\).

Step 3: Solving the system of equations

Next, solve the system of equations. First, solve \(\frac{\partial L}{\partial λ} = 0\) as \(y = \frac{15-2x}{4}\). Substitute this into \(\frac{x}{\sqrt{x^{2}+y^{2}}} - 2λ = 0\) to solve for \(λ\). After finding \(λ\), substitute it back to solve for \(x\), then substitute \(x\) into the constraint equation to solve for \(y\).

Step 4: Finding the maximum value

Finally, substitute the values of \(x\) and \(y\) obtained in the previous step into \(f(x, y)=\sqrt{x^{2}+y^{2}}\) to find the maximum value of the function.

Key Concepts

Objective Function

An objective function, in the context of optimization, is the function you want to maximize or minimize. In this exercise, it refers to the function of two variables, \( f(x, y) = \sqrt{x^{2}+y^{2}} \). This function represents the distance from the origin, a common formulation when the goal is to find extremities such as maximum distances.
This is typical in constraint optimization problems where an objective is defined subject to certain limitations or constraints. It's important to clearly understand the goal of the objective function, which in this example is maximizing the distance.

• Objective functions are central to optimization problems.
• They serve as the primary measure of success or goal.
• Formulated from the problem context, they represent what you're seeking to achieve.

To tackle the problem, one needs to express the objective function in terms of constraints, which often involves mathematical rearranging and substitutions.

Constraint Optimization

Constraint optimization is the process of finding the optimal solution, subject to certain conditions or restrictions, known as constraints. In this exercise, the constraint is given by the equation \(2x + 4y - 15 = 0\). This equation must hold true for any potential solutions for \(x\) and \(y\). By using the method of Lagrange multipliers, you can incorporate the constraints directly into the optimization process. This is done by defining a new function known as the Lagrangian.

• A constraint affects the feasible set of potential solutions.
• Lagrange multipliers are introduced to ensure the constraints are satisfied at optimal solutions.
• The solution is optimal when it satisfies both the derivative conditions and the constraint.

In essence, constraint optimization balances achieving the best possible outcome without violating given restrictions, ensuring practical and feasible solutions in real-life applications.

Partial Derivatives

Partial derivatives are a tool used to understand how a function changes as each of its variables changes individually, holding others constant. They play a critical role in optimization, particularly in problems involving multiple variables, like our exercise.
In the Lagrangian formulation, we derive the partial derivatives with respect to all components \( x \), \( y \), and the Lagrange multiplier \( \lambda \). Let's look at each step:

• \( \frac{\partial L}{\partial x} \) and \( \frac{\partial L}{\partial y} \) helps find the rate of change of \( L \) concerning \( x \) and \( y \).
• This results in a system of equations when the partial derivatives are set to zero, suggesting points where extremum might occur.
• \( \frac{\partial L}{\partial \lambda} \) helps ensure the constraint equation is satisfied.

For optimization via Lagrange multipliers, the system of equations from these derivatives needs solving to identify solutions that satisfy both the objective and the constraints adequately.