Problem 358 - 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. $$ f(x, y)=x^{2} y ; x^{2}+2 y^{2}=6 $$

Step-by-Step Solution

Answer

Maximum value: 4, Minimum value: -4.

Step 1: Understand the Problem

We need to find the maximum and minimum values of the function \( f(x, y) = x^2 y \) subject to the constraint \( g(x, y) = x^2 + 2y^2 = 6 \). This can be done using Lagrange multipliers.

Step 2: Define the Lagrangian Function

Define the Lagrangian function \( \mathcal{L}(x, y, \lambda) = x^2 y + \lambda (6 - x^2 - 2y^2) \), where \( \lambda \) is the Lagrange multiplier.

Step 3: Find the Partial Derivatives

Compute the partial derivatives:1. \( \frac{\partial \mathcal{L}}{\partial x} = 2xy - 2\lambda x = 0 \).2. \( \frac{\partial \mathcal{L}}{\partial y} = x^2 - 4\lambda y = 0 \).3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 6 - x^2 - 2y^2 = 0 \).

Step 4: Solve the System of Equations

Solve the system of equations:1. \( 2xy - 2\lambda x = 0 \) implies \( 2x(y - \lambda) = 0 \). Hence, \( x = 0 \) or \( y = \lambda \).2. \( x^2 = 4\lambda y \). Substituting \( y = \lambda \) gives \( x^2 = 4y^2 \).3. Use \( x^2 + 2y^2 = 6 \) with \( x = 0 \) or \( x^2 = 4y^2 \) to find solutions.

Step 5: Consider Case \( x = 0 \)

When \( x = 0 \), the constraint becomes \( 2y^2 = 6 \) leading to \( y = \pm \sqrt{3} \). Thus, the points are \((0, \sqrt{3})\) and \((0, -\sqrt{3})\).

Step 6: Consider Case \( x^2 = 4y^2 \)

Solving \( x^2 = 4y^2 \) and \( x^2 + 2y^2 = 6 \), substitute \( x^2 = 4y^2 \) into \( 6 - 6y^2 = 0 \) to find \( y = \pm 1 \). This gives the points \( (\pm 2, 1) \) and \( (\pm 2, -1) \).

Step 7: Evaluate \( f(x, y) \) at Found Points

Calculate \( f(x, y) = x^2 y \) at all critical points:- \( f(0, \sqrt{3}) = 0 \).- \( f(0, -\sqrt{3}) = 0 \).- \( f(2, 1) = 4 \).- \( f(2, -1) = -4 \).- \( f(-2, 1) = 4 \).- \( f(-2, -1) = -4 \).

Step 8: Identify Maximum and Minimum Values

The maximum value of \( f(x, y) \) is \( 4 \) at the points \( (2, 1) \) and \( (-2, 1) \). The minimum value is \( -4 \) at the points \( (2, -1) \) and \( (-2, -1) \).

Key Concepts

Constraint Optimization

Constraint optimization is a method used to find the extremum (maximum or minimum) of a function, subject to certain limitations or restrictions, known as constraints. In our exercise, we are particularly interested in optimizing the function \( f(x, y) = x^2 y \) under the given constraint \( g(x, y) = x^2 + 2y^2 = 6 \). This type of problem is quite common in real-world applications, where resources or conditions impose limits on what solutions are feasible.

To tackle these problems effectively, the method of Lagrange multipliers is employed. This approach involves transforming the original problem by introducing a new function known as the Lagrangian. This function incorporates both the objective (the function we wish to optimize) and the constraint, providing a unified way to solve for all possible solutions concurrently.

Partial Derivatives

Partial derivatives are an essential concept in multivariable calculus, helping us understand how functions change with respect to each variable while keeping the other variables constant. In the context of the Lagrange multiplier method, partial derivatives are used to derive the critical equations necessary for optimizing the Lagrangian.

For our exercise, we compute the partial derivatives of the Lagrangian \( \mathcal{L}(x, y, \lambda) = x^2 y + \lambda (6 - x^2 - 2y^2) \). Here are the ones we need:

• \( \frac{\partial \mathcal{L}}{\partial x} = 2xy - 2\lambda x = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial y} = x^2 - 4\lambda y = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = 6 - x^2 - 2y^2 = 0 \)

These derivatives help us set up a system of equations to solve for the variables \( x, y, \) and the Lagrange multiplier \( \lambda \). By doing this, we identify the points where the function could potentially reach a maximum or minimum value given the constraint.

Critical Points

Once you have the partial derivatives, the next goal is to find the critical points. Critical points are essentially solutions to the derivative equations that could represent maximum, minimum, or saddle points. These points are where the gradient (a vector of partial derivatives) is zero or undefined, indicating that there is no straightforward increase or decrease in the function at these locations.

For our exercise, solving the derivatives gives us two main cases to consider:

• When \( x = 0 \).
• When \( x^2 = 4y^2 \).

In the first case, solving the constraint \( 2y^2 = 6 \) helps us find the possible critical points like \((0, \sqrt{3})\) and \((0, -\sqrt{3})\). In the second case, by linking \( x^2 = 4y^2 \) and the constraint \( x^2 + 2y^2 = 6 \), we identify additional critical points like \((\pm 2, 1)\) and \((\pm 2, -1)\). These are potential candidates where the function could achieve its highest or lowest values, given the restriction in place.

Maximum and Minimum Values

Finally, to determine the maximum and minimum values of the function, evaluate it at all the critical points found earlier. This involves plugging each point back into the original function \( f(x, y) = x^2 y \) and calculating the resulting values.

In our exercise, evaluating the function at the critical points yields:

• \( f(0, \sqrt{3}) = 0 \)
• \( f(0, -\sqrt{3}) = 0 \)
• \( f(2, 1) = 4 \)
• \( f(2, -1) = -4 \)
• \( f(-2, 1) = 4 \)
• \( f(-2, -1) = -4 \)

The maximum value of the function, which is \( 4 \), is reached at points \( (2, 1) \) and \( (-2, 1) \). The minimum value of \( -4 \) occurs at points \( (2, -1) \) and \( (-2, -1) \). These values tell us how the function behaves under the restraint imposed by the constraint, depicting both its limits and potential beyond the unconstrained scenario.