Problem 10 - College Algebra and Calculus: An Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \text { Minimize } f(x, y)=\sqrt{x^{2}+y^{2}} \quad 2 x+4 y-15=0 $$

Step-by-Step Solution

Answer

Begin by constructing a Lagrangian function, compute and set the partial derivatives to zero, solve equations obtained for \(x\), \(y\), and \(\lambda\) and finally substitute the obtained values in the original function to get the minimum value.

Step 1: Set up the Lagrange function

The Lagrange function is set by adding the constraint equation multiplied by a new variable, \(\lambda\), to the original function. The new Lagrange function is \(F(x,y,\lambda)=\sqrt{x^{2}+y^{2}}+\lambda (2x+4y-15)\).

Step 2: Find the partial derivatives

The partial derivatives of the function wrt variables \(x\), \(y\), and \(\lambda\) are computed as follows: \(\frac{\partial F}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}}+2\lambda\), \(\frac{\partial F}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}}+4\lambda\), and \(\frac{\partial F}{\partial \lambda}=2x+4y-15\). Then set each derivative equal to zero.

Step 3: Solve the equations

Solving these equations will give the values of \(x\), \(y\), and \(\lambda\) that will minimize \(f(x, y)\). Using the equations \(\frac{x}{\sqrt{x^{2}+y^{2}}}+2\lambda=0\), \(\frac{y}{\sqrt{x^{2}+y^{2}}}+4\lambda=0\), and \(2x+4y-15=0\), one can solve for \(x\), \(y\), and \(\lambda\).

Step 4: Substitute the values of x and y into the original function

After finding the values of \(x\), \(y\), and \(\lambda\), these values need to be substituted back in to the original \(f(x, y)\) equation. With this process, the minimum value of \(f(x, y)\) can be computed.

Key Concepts

Optimization

Optimization is a mathematical process used to find the best solution from a set of choices. In this exercise, we aim to minimize a function using an optimization technique known as Lagrange multipliers. Here, we strive to make the distance, represented by the function \( f(x, y) = \sqrt{x^2 + y^2} \), as small as possible, subject to a specific constraint \( 2x + 4y - 15 = 0 \). Finding this minimum efficiently is critical, as it allows us to determine optimal solutions in various real-world situations such as economics, engineering, and logistics.

• The goal is to find points where the function attains minima or maxima, given a limitation or constraint.
• By considering constraints, we only look at feasible solutions that satisfy certain conditions.

The use of Lagrange multipliers is particularly helpful because it incorporates the constraint directly into the optimization process, allowing us to simultaneously address multiple variables and constraints.

Partial Derivatives

Partial derivatives are a fundamental concept in calculus used to analyze functions with more than one variable. When calculating partial derivatives, we determine how a function changes as each variable changes individually while keeping the other variables constant.

• For the function \( F(x, y, \lambda) = \sqrt{x^2 + y^2} + \lambda(2x + 4y - 15) \), partial derivatives are computed with respect to each variable \( x \), \( y \), and \( \lambda \).
• This process reveals how sensitive the function is to changes in each parameter under the constraint conditions.

Finding partial derivatives enables us to extract important information about the rate of change in different directions, an essential step in solving optimization problems with multiple variables. By setting these derivatives to zero, we identify critical points where the function could be optimized.

Constraint Equations

Constraint equations are additional conditions that solutions to a problem must satisfy. These constraints limit the possible values variables can take and are essential in guiding the optimization process. In the problem, the constraint is \( 2x + 4y - 15 = 0 \). This means that any solution for \( x \) and \( y \) must satisfy this equation.

• The constraint represents a boundary or a rule governing the relationship between variables.
• Integrating the constraint with the primary function via Lagrange multipliers allows us to find solutions that respect these boundaries.

Understanding constraint equations allows us to narrow down the search for optimal solutions significantly by excluding infeasible ones that don't meet the required conditions.

Calculus

Calculus provides the tools needed to analyze and solve problems involving changes. It is particularly powerful in optimization because it gives us methods like differentiation and integration.

• By using calculus, we compute partial derivatives and identify critical points where a function might attain a minimum or maximum value.
• Through processes like differentiation, calculus helps to identify solutions that satisfy both the main function and its constraints.

In this exercise, calculus plays a central role in applying Lagrange multipliers, ensuring that we effectively integrate the constraint into the optimization calculation. Calculus, therefore, provides the mathematical foundation that allows us to explore changes and interactions between variables within constrained environments.