Problem 3 - Applied Calculus

Question

Use Lagrange multipliers to find the maximum or minimum values of \(f(x, y)\) subject to the constraint. $$f(x, y)=x y, \quad 5 x+2 y=100$$

Step-by-Step Solution

Answer

Maximum value is 250.

Step 1: Understanding the problem

We need to find the maximum or minimum values of the function \( f(x, y) = xy \) subject to the constraint \( 5x + 2y = 100 \). We will use the method of Lagrange multipliers to solve this problem.

Step 2: Setting up the Lagrange function

The Lagrange function is defined as \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda(g(x, y) - c) \), where \( g(x, y) = 5x + 2y \) is the constraint and \( c = 100 \). We have: \( \mathcal{L}(x, y, \lambda) = xy + \lambda(5x + 2y - 100) \).

Step 3: Finding the partial derivatives

Compute the partial derivatives of \( \mathcal{L}(x, y, \lambda) \) with respect to \( x, y, \) and \( \lambda \). 1. \( \frac{\partial \mathcal{L}}{\partial x} = y + 5\lambda \) 2. \( \frac{\partial \mathcal{L}}{\partial y} = x + 2\lambda \) 3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 5x + 2y - 100 \)

Step 4: Solving the system of equations

Set the partial derivatives equal to zero and solve the system of equations: 1. \( y + 5\lambda = 0 \)2. \( x + 2\lambda = 0 \)3. \( 5x + 2y = 100 \).

Step 5: Solve for \( \lambda \)

From equations 1 and 2: \( y = -5\lambda \) and \( x = -2\lambda \).

Step 6: Substitute \( y \) and \( x \) into the constraint

Substitute \( y = -5\lambda \) and \( x = -2\lambda \) into \( 5x + 2y = 100 \) to find \( \lambda \):\[ 5(-2\lambda) + 2(-5\lambda) = 100 \] \[ -10\lambda - 10\lambda = 100 \] \[ -20\lambda = 100 \] \( \lambda = -5 \).

Step 7: Find \( x \) and \( y \) values

Substitute \( \lambda = -5 \) back to expressions for \( x \) and \( y \):\( x = -2(-5) = 10 \) \( y = -5(-5) = 25 \).

Step 8: Calculate \( f(x, y) \)

Now compute \( f(x, y) = xy \):\( f(10, 25) = 10 \times 25 = 250 \).

Step 9: Conclusion

The maximum value of the function \( f(x, y) = xy \) given the constraint \( 5x + 2y = 100 \) is 250.

Key Concepts

Constrained Optimization

Constrained optimization is a fundamental concept in mathematical optimization. It's used when you need to find the best solution (like maximum or minimum values) out of a set of possibilities that satisfy certain conditions, termed constraints.
In the context of this problem, the goal is to optimize the function \( f(x, y) = xy \) under the given constraint \( 5x + 2y = 100 \).

• The optimization is not "free," meaning we can't just pick any values for \( x \) and \( y \). They have to fit the constraint.
• Lagrange multipliers help us find optimal solutions without needing to directly solve the constraint, providing a strategy that's widely applicable.

Constrained optimization problems are everywhere in real life, like maximizing profits while respecting resource limits.

Partial Derivatives

Partial derivatives are a crucial tool in optimization problems where functions have several variables. They measure how a function changes as one variable changes, holding the other variables constant.
In this exercise, we use partial derivatives with respect to \( x \), \( y \), and \( \lambda \) to form equations that lead us to the solution.

• For \( \mathcal{L}(x, y, \lambda) \), we compute:
  • \( \frac{\partial \mathcal{L}}{\partial x} = y + 5\lambda \)
  • \( \frac{\partial \mathcal{L}}{\partial y} = x + 2\lambda \)
  • \( \frac{\partial \mathcal{L}}{\partial \lambda} = 5x + 2y - 100 \)
• Setting these derivatives to zero helps us find critical points by considering the rates of change.

Partial derivatives are essential in understanding how changes affect a system, and they are a stepping stone to solving the equations formed in constrained optimization.

System of Equations

A system of equations involves multiple equations that share common variables. Solving it gives us values that satisfy all conditions simultaneously.
In this problem, the partial derivatives form a system:

• \( y + 5\lambda = 0 \)
• \( x + 2\lambda = 0 \)
• \( 5x + 2y = 100 \)

This system must be solved together to keep the constraint satisfied while optimizing.

• Each equation contributes to a unique insight or condition that the variables must meet.
• Simplifying and solving these equations sequentially helps find the values of \( x \), \( y \), and \( \lambda \) that work for both the function and the constraint.

Systems of equations are a common way to manage multiple dependencies and conditions within optimization problems.

Mathematical Optimization

Mathematical optimization is the overarching field that involves finding the best solution from a set of possible choices. It often involves maximizing or minimizing a function within a defined domain.
In this example, we use Lagrange multipliers for solving an optimization problem, which is a technique from this field.

• Optimization is used to make efficient decisions and is applied in various industries like finance, engineering, and logistics.
• It's about finding solutions not just via straightforward calculations, but using strategies like Lagrange multipliers for more complex scenarios.

Understanding mathematical optimization helps frame real-world problems into solvable mathematical models, improving efficiency and outcomes in various applications.