Problem 361 - Calculus Volume 3

Question

Use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints. $$ f(x, y)=4 x^{3}+y^{2} ; 2 x^{2}+y^{2}=1 $$

Step-by-Step Solution

Answer

Maximum is 1; minimum is \(-\sqrt{2}/2\).

Step 1: Identify the function and constraint

We are given the function: \( f(x, y) = 4x^3 + y^2 \) that needs to be optimized. The constraint is \( 2x^2 + y^2 = 1 \). The goal is to find the maximum and minimum values of \( f(x, y) \) subject to this constraint using the method of Lagrange multipliers.

Step 2: Set up Lagrangian

Introduce a Lagrange multiplier \( \lambda \) and set up the Lagrangian function: \[ \mathcal{L}(x, y, \lambda) = 4x^3 + y^2 + \lambda(1 - 2x^2 - y^2) \] Here, the multiplier \( \lambda \) is used to incorporate the constraint into the optimization process.

Step 3: Compute partial derivatives

Compute the partial derivatives of the Lagrangian function with respect to \( x \), \( y \), and \( \lambda \), and set each derivative equal to zero:1. \( \frac{\partial \mathcal{L}}{\partial x} = 12x^2 - 4\lambda x = 0 \)2. \( \frac{\partial \mathcal{L}}{\partial y} = 2y - 2\lambda y = 0 \)3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 1 - 2x^2 - y^2 = 0 \) The solutions to these equations will give us potential extrema.

Step 4: Solve Lagrange equations

From \( 2y(1-\lambda) = 0 \), we find either \( y = 0 \) or \( \lambda = 1 \).- If \( y = 0 \), from the constraint: \( 2x^2 = 1 \) so \( x = \pm \frac{1}{\sqrt{2}} \).- If \( \lambda = 1 \), use \( 4x(3x-\lambda) = 0 \) to find \( x = 0 \) or \( \lambda = 3x \).When \( x = 0 \), from the constraint: \( y^2 = 1 \) so \( y = \pm 1 \).

Step 5: Evaluate function at critical points

Evaluate \( f(x, y) = 4x^3 + y^2 \) at each critical point:- When \( x = \frac{1}{\sqrt{2}} \) and \( y = 0 \): \( f(x, y) = 4\left(\frac{1}{\sqrt{2}}\right)^3 = \sqrt{2}/2 \).- When \( x = -\frac{1}{\sqrt{2}} \) and \( y = 0 \): \( f(x, y) = -\sqrt{2}/2 \).- When \( x = 0 \) and \( y = 1 \): \( f(x, y) = 1 \).- When \( x = 0 \) and \( y = -1 \): \( f(x, y) = 1 \).

Step 6: Determine maximum and minimum values

The function values at the critical points are \( \pm \sqrt{2}/2 \) and \( 1 \). The maximum value is \( f(x, y) = 1 \) and the minimum is \( f(x, y) = -\sqrt{2}/2 \).

Key Concepts

Optimization

Optimization involves finding the best solution from a set of possible options. In mathematics, it usually refers to finding the maximum or minimum values of a function. This is crucial in many fields, as it helps identify the most efficient solutions. In our exercise, we optimize the function \( f(x, y) = 4x^3 + y^2 \). The goal is to determine where this function reaches its highest and lowest points, known as the extrema, while considering a given constraint.

Optimization can be applied in

• Economics: to maximize profits
• Physics: to minimize energy use
• Engineering: to optimize designs

Each of these areas relies on optimizing functions to ensure efficiency and effectiveness. The concept often involves calculus, as finding maxima and minima typically requires derivatives.

Constraints

Constraints are conditions that must be satisfied for a solution to be considered valid. They limit the possible values the variables can take. In our example, the constraint is given by the equation \(2x^2 + y^2 = 1\). This constraint forms an ellipse, and our task is to optimize the function \(f(x, y) = 4x^3 + y^2\) within this boundary.

Constraints can come in different forms, such as

• Inequalities, like \(x + y \leq 5\)
• Equations, like \(z = x^2 + y^2\)

These restrictions are integral to the problem, as they define where the function can be evaluated. Without constraints, optimization would result in a different solution as it would span the entire domain of the function.

Calculus

Calculus is the branch of mathematics that studies change and motion. It is indispensable in solving problems involving optimization, especially in continuous domains. By using derivatives, calculus helps us find the slope of functions at given points, allowing the determination of where these slopes are zero—points where extrema likely occur.

In our use of Lagrange multipliers, calculus helps through:

• Finding partial derivatives with respect to variables in a function
• Setting up and solving systems of equations to find stationary points

Without calculus, solving the optimization problem with constraints would be significantly more challenging, as these derivatives are necessary for locating potential maxima and minima.

Extrema

Extrema are the maximum or minimum values that a function can achieve. In optimization, finding these values is often the primary goal. Our aim is to discover where \( f(x, y) = 4x^3 + y^2 \) is at its highest or lowest, given the constraint \(2x^2 + y^2 = 1\).

For a function subject to constraints, extrema can be found using Lagrange multipliers:

• The Lagrangian is used to incorporate constraints into the function
• Critical points are evaluated to find the highest and lowest values

These calculations yield the function's extrema, which in this problem were \(1\) as the maximum and \(-\sqrt{2}/2\) as the minimum. Extrema provide valuable insight into the behavior of a function within the given limits, and are a key focus in various optimization problems.