Question

Find the minimum of \(f(x, y)=x^{2}+y^{2}\) subject to the constraint \(g(x, y)=x y-3=0\)

Short answer

The minimum value is 6 at points \((\sqrt{3}, \sqrt{3})\) and \((-\sqrt{3}, -\sqrt{3})\).

Step-by-step solution

Identify the Lagrange Function

To solve this problem, we'll use the method of Lagrange multipliers. First, identify the Lagrangian: \ \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda \cdot g(x, y) \), where \( f(x, y) = x^2 + y^2 \) and \( g(x, y) = xy - 3 \). So, \( \mathcal{L}(x, y, \lambda) = x^2 + y^2 + \lambda(xy - 3) \).

Calculate Partial Derivatives

Next, calculate the partial derivatives of \( \mathcal{L} \) with respect to \( x, y, \) and \( \lambda \): (1) \( \frac{\partial \mathcal{L}}{\partial x} = 2x + \lambda y \)(2) \( \frac{\partial \mathcal{L}}{\partial y} = 2y + \lambda x \)(3) \( \frac{\partial \mathcal{L}}{\partial \lambda} = xy - 3 \)

Set Partial Derivatives to Zero

Set each of these partial derivatives to zero to find the equations:(1) \( 2x + \lambda y = 0 \)(2) \( 2y + \lambda x = 0 \)(3) \( xy - 3 = 0 \)

Solve the System of Equations

Solve the system of equations:From (1), \( \lambda y = -2x \) \( \Rightarrow \lambda = -\frac{2x}{y} \).From (2), \( \lambda x = -2y \) \( \Rightarrow \lambda = -\frac{2y}{x} \).Equating the two expressions for \( \lambda \): \(-\frac{2x}{y} = -\frac{2y}{x} \), which simplifies to \( x^2 = y^2 \) or \( x = \pm y \).

Apply the Constraint

Use the constraint \( xy = 3 \):Case 1: \( x = y \), then \( x^2 = 3 \) leading to \( x = \sqrt{3} \) or \( x = -\sqrt{3} \).Case 2: \( x = -y \), then \( -x^2 = 3 \) which is not possible as \( x^2 \geq 0 \).

Evaluate the Function

Evaluate the function \( f(x, y) = x^2 + y^2 \) at the points found:For \( (x, y) = (\sqrt{3}, \sqrt{3}) \) and \( (x, y) = (-\sqrt{3}, -\sqrt{3}) \), calculate: \( f(\sqrt{3}, \sqrt{3}) = (\sqrt{3})^2 + (\sqrt{3})^2 = 6 \).Both points yield the same minimum value of 6 for \( f(x, y) \).

Key concepts

Constrained Optimization

Constrained optimization is a core concept in calculus where we locate the optimal values of a function subject to restrictions, or constraints. Instead of simply finding maxima or minima, we identify these points while adhering to certain conditions. In this exercise, we're asked to find the minimum value of the function \( f(x, y) = x^2 + y^2 \), subject to the constraint \( g(x, y) = xy - 3 = 0 \). This means the solution must satisfy the constraint, ensuring that while \( f(x, y) \) is optimized, the product of \( x \) and \( y \) is always 3.

The method uses Lagrange multipliers, introducing a new variable, \( \lambda \), that "weighs" the constraint alongside the function we want to optimize. This approach essentially combines the original problem with the constraint into a single function, the Lagrangian. It mathematically expresses how the constraint impacts \( f(x, y) \), enabling an efficient search for the optimization points.

Partial Derivatives

Partial derivatives are the building blocks of calculus involving functions with multiple variables. By taking the derivative with respect to one variable while keeping others constant, we understand how changes in that single variable affect the whole function. In the context of optimization, partial derivatives allow us to explore the direction and rate of change of a function's slope for each variable individually.

In this exercise, we compute the partial derivatives of the Lagrangian with respect to \( x, y, \) and \( \lambda \). It provides the necessary equations to identify the critical points where our function could achieve a minimum or maximum under the stated constraint. The derivatives \( \frac{\partial \mathcal{L}}{\partial x} = 2x + \lambda y \), \( \frac{\partial \mathcal{L}}{\partial y} = 2y + \lambda x \), and \( \frac{\partial \mathcal{L}}{\partial \lambda} = xy - 3 \) set the stage for solving the equations to find these critical points.

System of Equations

After computing partial derivatives, the next step involves solving the system of equations that results when these derivatives are set to zero. Solving a system of equations is crucial in finding the values where our constrained optimization achieves potential extremum values.

By setting \( 2x + \lambda y = 0 \), \( 2y + \lambda x = 0 \), and \( xy - 3 = 0 \), we have three equations that contain the variables \( x, y, \) and \( \lambda \). Solving these provides possible solutions for the optimization problem. Where equations (1) and (2) derive the critical condition \( x = \pm y \), and equation (3) reaffirms the constraint condition. The interplay of these equations helps in isolating variables and determining viable solutions that also adhere to \( xy = 3 \).

Function Evaluation

Function evaluation is the final step in constrained optimization where we substitute back the obtained solutions to find the minimum or maximum values of the original function. This ensures that the values indeed meet the problem's requirements and how optimal they truly are.

For this problem, we evaluated the function \( f(x, y) = x^2 + y^2 \) at solutions \( (x, y) = (\sqrt{3}, \sqrt{3}) \) and \( (x, y) = (-\sqrt{3}, -\sqrt{3}) \). Since these points satisfy the condition \( xy = 3 \) and no other pairs fulfill this, calculating \( f(\sqrt{3}, \sqrt{3}) \) gives us \( 6 \), which is the minimum value within the constraints. This demonstrates the final verification of the optimization process under given conditions, confirming the accuracy and validity of our solutions.