Problem 2 - Applied Calculus

Question

Use Lagrange multipliers to find the maximum or minimum values of \(f(x, y)\) subject to the constraint. $$f(x, y)=x^{2}+4 x y, \quad x+y=100$$

Step-by-Step Solution

Answer

The maximum value is \( \frac{40000}{9} \) when \( x = \frac{200}{3} \) and \( y = \frac{100}{3} \).

Step 1: Introduction of Lagrange Multiplier

To find the extrema of the function \( f(x, y) = x^2 + 4xy \) subject to the constraint \( x + y = 100 \), we apply the method of Lagrange multipliers. We introduce a Lagrange multiplier \( \lambda \) and form the Lagrangian function: \( \mathcal{L}(x, y, \lambda) = x^2 + 4xy + \lambda(100 - x - y) \).

Step 2: Derivatives of the Lagrangian

Compute the partial derivatives of the Lagrangian \( \mathcal{L}(x, y, \lambda) \) with respect to \( x \), \( y \), and \( \lambda \):1. \( \frac{\partial \mathcal{L}}{\partial x} = 2x + 4y - \lambda \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 4x - \lambda \)3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 100 - x - y \)

Step 3: Setting Derivatives to Zero

Set each partial derivative to zero to find the critical points:1. \( 2x + 4y - \lambda = 0 \)2. \( 4x - \lambda = 0 \)3. \( 100 - x - y = 0 \)

Step 4: Solving the System of Equations

From equation 2, \( \lambda = 4x \). Substitute \( 4x \) for \( \lambda \) in equation 1:\( 2x + 4y - 4x = 0 \) simplifies to \( 4y = 2x \) or \( y = \frac{x}{2} \).Substitute \( y = \frac{x}{2} \) into the constraint \( 100 - x - y = 0 \):\( 100 - x - \frac{x}{2} = 0 \) simplifies to \( 100 = \frac{3x}{2} \).Solving for \( x \), we get \( x = \frac{200}{3} \).Substitute \( x = \frac{200}{3} \) back to find \( y \): \( y = \frac{\frac{200}{3}}{2} = \frac{100}{3} \).

Step 5: Verification and Conclusion

Substitute \( x = \frac{200}{3} \) and \( y = \frac{100}{3} \) into the constraint to verify:Constraint check: \( x + y = \frac{200}{3} + \frac{100}{3} = 100 \), which satisfies the constraint. Thus, the solution is valid.Substitute back into the function to find the function value: \( f\left( \frac{200}{3}, \frac{100}{3} \right) = \left( \frac{200}{3} \right)^2 + 4 \cdot \frac{200}{3} \cdot \frac{100}{3} = \frac{40000}{9} \).

Key Concepts

Constrained Optimization

Lagrange multipliers are a powerful technique used in constrained optimization problems. The goal is to find the maximum or minimum of a function given a constraint. This is useful in situations where resources are limited or conditions must be met. For our exercise, the function \( f(x, y) = x^2 + 4xy \) needs to be optimized under the constraint \( x + y = 100 \). We cannot simply find extrema as usual because of the constraint.
Instead, we introduce a new variable, called the Lagrange multiplier, denoted by \( \lambda \). This transforms the problem into finding the extrema of a new function, the Lagrangian: \( \mathcal{L}(x, y, \lambda) = x^2 + 4xy + \lambda(100 - x - y) \). The magic of Lagrange multipliers is that they help find critical points where the constraint is satisfied.

Partial Derivatives

Partial derivatives allow us to investigate how a function changes as each variable changes independently. For a given multivariable function, partial derivatives are essential.
In the Lagrange multiplier method, we take partial derivatives of the Lagrangian function with respect to \( x \), \( y \), and \( \lambda \).

• For \( x \): the derivative is \( \frac{\partial \mathcal{L}}{\partial x} = 2x + 4y - \lambda \).
• For \( y \): the derivative is \( \frac{\partial \mathcal{L}}{\partial y} = 4x - \lambda \).
• For \( \lambda \): the derivative is \( \frac{\partial \mathcal{L}}{\partial \lambda} = 100 - x - y \).

These derivatives tell us the rate of change of the Lagrangian with respect to each variable, which is crucial for finding where the function reaches its extremum under the constraint.

Critical Points

Critical points are where the function's rate of change is zero. For constrained optimization, these are the values of variables that might yield the maximum or minimum of the function.
To find critical points, we set each of the partial derivatives to zero:

• \( 2x + 4y - \lambda = 0 \)
• \( 4x - \lambda = 0 \)
• \( 100 - x - y = 0 \)

This creates a system of equations. Solving these equations finds possible values of \( x \), \( y \), and \( \lambda \). Solving them step-by-step helps locate the points that potentially offer the best or worst value of the original function under the constraint.

Extrema of Multivariable Functions

Multivariable functions like \( f(x, y) = x^2 + 4xy \) can have maximum or minimum values, known as extrema. Finding these is complex when more than one variable is involved, especially when a constraint is applied.
Let's substitute our critical points into the function to verify whether they are indeed extrema. In our example, solving the equations and constraints gives us \( x = \frac{200}{3} \) and \( y = \frac{100}{3} \). Verifying the constraint, \( x + y = 100 \), shows that the solutions are consistent.
Finally, substitute these values back into the original function to calculate its value at these points. This will tell us whether we've found a maximum or minimum, as exemplified by \( f\left( \frac{200}{3}, \frac{100}{3} \right) = \frac{40000}{9} \), representing the extremum of the function under the given condition.