Problem 360 - 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. $$ f(x, y)=x y ; 4 x^{2}+8 y^{2}=16 $$

Step-by-Step Solution

Answer

Maximum value is 2, and minimum value is -2.

Step 1: Identify the Objective Function and Constraint

The objective function is the one we need to optimize, which in this case is given by \[ f(x, y) = xy \] The constraint is the equation related to the condition that needs to be satisfied. In this case, the constraint is given by \[ 4x^2 + 8y^2 = 16 \]

Step 2: Write the Lagrangian

To incorporate the constraint, we form the Lagrangian function, \[ \mathcal{L}(x, y, \lambda) = xy - \lambda(4x^2 + 8y^2 - 16) \]where \( \lambda \) is the Lagrange multiplier. This function will help us find critical points that satisfy both the objective function and the constraint.

Step 3: Set Up the System of Equations

We need to find the partial derivatives of the Lagrangian and set them to zero:1. \[ \frac{\partial \mathcal{L}}{\partial x} = y - \lambda(8x) = 0 \]2. \[ \frac{\partial \mathcal{L}}{\partial y} = x - \lambda(16y) = 0 \]3. \[ \frac{\partial \mathcal{L}}{\partial \lambda} = -(4x^2 + 8y^2 - 16) = 0 \]These equations ensure that both the objective function is optimized and the constraint is satisfied.

Step 4: Solve the System of Equations

First, solve the equations \( y = 8\lambda x \) and \( x = 16\lambda y \) simultaneously. By substitution or elimination, we find \( y = 2x \) or \( x = -2y \). Substitute this into the constraint:- If \( y = 2x \), substitute into constraint \[ 4x^2 + 8(2x)^2 = 16 \] Solving, we get \( x = \pm 1 \), \( y = \pm 2 \).- If \( x = -2y \), substitute into constraint \[ 4(-2y)^2 + 8y^2 = 16 \]Solving, we get \( x = \pm 2 \), \( y = \mp 1 \).

Step 5: Evaluate the Objective Function at Critical Points

Plug the critical points back into the objective function:- At \((x, y) = (1, 2)\) or \((x, y) = (-1, -2)\), \[ f(1, 2) = 1 \times 2 = 2 \] \[ f(-1, -2) = (-1) \times (-2) = 2 \] - At \((x, y) = (2, -1)\) or \((x, y) = (-2, 1)\), \[ f(2, -1) = 2 \times (-1) = -2 \] \[ f(-2, 1) = (-2) \times 1 = -2 \]the maximum value of \(f(x, y)\) is 2, and the minimum value is -2.

Key Concepts

Multivariable Calculus

Multivariable calculus involves the extension of calculus to functions of more than one variable. In this context, we're working with functions like \( f(x, y) = xy \), which depend on both \( x \) and \( y \).
To optimize such functions, we analyze how they change when we vary these variables. By using partial derivatives, we determine the rate of change in each variable separately, holding the others constant.
The derivative with respect to \( x \), for instance, tells us how much \( f(x, y) \) changes along the \( x \)-axis. Similarly, the derivative with respect to \( y \) shows changes along the \( y \)-axis.
Partial derivatives are foundational in calculating gradients and understanding how a multi-variable function behaves in its domain. These concepts are pivotal when employing Lagrange multipliers for obtaining optimal solutions under constraints.

Constraint Optimization

Constraint optimization refers to finding the maximum or minimum of a function while adhering to specific conditions or constraints.
In our example, the goal was to optimize the function \( f(x, y) = xy \) while ensuring the constraint \( 4x^2 + 8y^2 = 16 \) holds. This often involves setting up an auxiliary equation called the Lagrangian.

• The Lagrangian incorporates both the objective function and its constraints, providing a single equation to solve.
• Using the Lagrangian reformulates the problem to find points where the gradient of the objective function aligns with the gradient of the constraint.
• Solving the corresponding system of equations yields critical points that respect both the function and its constraint, offering feasible solutions for optimization.

The method of Lagrange multipliers is crucial for such types of problems, simplifying the process of finding appropriate solutions within imposed boundaries.

Critical Points

Critical points refer to the values of variables where the derivative of a function equals zero or where it's undefined.
These points indicate where a function displays a local maximum, minimum, or saddle point. In multivariable calculus, finding these points involves setting partial derivatives of the Lagrangian to zero.

• The solutions to these equations lead us to candidate points for maximum or minimum values, subjected to the constraint.
• Critical points in constraint optimization are also where the objective function tends to be tangent to the constraint surface.
• Once these critical points are identified, we evaluate the objective function at each point to determine which yields maximum and minimum values.

Identifying and analyzing critical points is a vital step in understanding the behavior of multivariable functions and optimizing them in relation to their constraints.