Problem 369 - Calculus Volume 3

Question

Minimize \(f(x, y)=x^{2}+y^{2}\) on the hyperbola \(x y=1\).

Step-by-Step Solution

Answer

The minimum value is 2 at points (1, 1) and (-1, -1).

Step 1: Identify the Objective and Constraint

In this problem, our goal is to minimize the function \( f(x, y) = x^2 + y^2 \) given the constraint \( xy = 1 \). This indicates that we are dealing with a constrained optimization problem where the constraint is a hyperbola.

Step 2: Set up the Lagrangian Function

For problems involving constraints, we use the method of Lagrange multipliers. We define the Lagrangian function as \( \mathcal{L}(x, y, \lambda) = x^2 + y^2 + \lambda(1 - xy) \), where \( \lambda \) is the Lagrange multiplier.

Step 3: Find the Partial Derivatives

We take the partial derivatives of the Lagrangian with respect to \(x\), \(y\), and \(\lambda\): \( \frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda y \), \( \frac{\partial \mathcal{L}}{\partial y} = 2y - \lambda x \), \( \frac{\partial \mathcal{L}}{\partial \lambda} = 1 - xy \).

Step 4: Solve the System of Equations

Set each of the partial derivatives equal to zero to form the system of equations: 1. \( 2x - \lambda y = 0 \) 2. \( 2y - \lambda x = 0 \) 3. \( 1 - xy = 0 \).Solve these equations simultaneously.

Step 5: Solve Derived Equations for \\(\lambda\\)\

From equations 1 and 2, derive the following: From (1), \( \lambda = \frac{2x}{y} \)From (2), \( \lambda = \frac{2y}{x} \)Equate these two expressions: \( \frac{2x}{y} = \frac{2y}{x} \)This simplifies to \( x^2 = y^2 \), hence \( x = y \) or \( x = -y \).

Step 6: Use the Constraint to Find Solutions

Recall the constraint \( xy = 1 \). Solve for \(x\) and \(y\) using both \(x = y\) and \(x = -y\):1. If \( x = y \), then \( x^2 = 1 \), giving \( x = 1 \) and \( y = 1 \) or \( x = -1 \) and \( y = -1 \).2. If \( x = -y \), substitute to get \( (x)(-x) = -x^2 = 1 \), which has no real solutions.

Step 7: Evaluate the Function at the Solutions

Evaluate \( f(x, y) = x^2 + y^2 \) at the feasible solutions:1. \((x, y) = (1, 1)\) gives \( f(x, y) = 1^2 + 1^2 = 2 \).2. \((x, y) = (-1, -1)\) also gives \( f(x, y) = (-1)^2 + (-1)^2 = 2 \).

Step 8: Conclusion

The minimum value of the function \( f(x, y) = x^2 + y^2 \) with the constraint \( xy = 1 \) is \(2\), achieved at both points \((1, 1)\) and \((-1, -1)\).

Key Concepts

Constrained Optimization

Constrained optimization is a mathematical method used when you want to find the minimum or maximum of a function within certain limits or constraints. In our exercise, the goal is to minimize the function \( f(x, y) = x^2 + y^2 \) under the condition that the product of \( x \) and \( y \) equals 1, which is our constraint. Constraints can define specific boundaries that solutions must adhere to, like a hyperbola in this case. By using constrained optimization techniques like Lagrange multipliers, we can find solutions that satisfy both the objective function and the given constraints simultaneously.

Hyperbola

A hyperbola is a type of curve on the graph that can be described by an equation like \( xy = c \), where \( c \) is a constant. In our case, the hyperbola is defined by \( xy = 1 \). This constraint creates a two-branched shape on a coordinate plane. Hyperbolas represent points where two quantities multiply to give a constant product. Thus, any valid solution for our optimization problem must lie on this hyperbola, meaning the points \((x, y)\) must satisfy that their product is always 1.

Partial Derivatives

To solve constrained optimization problems, it's useful to take partial derivatives. Partial derivatives are like regular derivatives, but they focus on how a function changes as one variable changes, keeping the others constant. For the Lagrangian function \( \mathcal{L}(x, y, \lambda) = x^2 + y^2 + \lambda(1 - xy) \), we take partial derivatives with respect to \( x \), \( y \), and \( \lambda \). This leads us to a system of equations. These derivatives help identify where the change in the Lagrangian is zero—a helpful clue in finding the minimum or maximum of our objective function subject to the constraint.

System of Equations

A system of equations is a set of equations with multiple variables that you solve together. In our step-by-step solution, once we took the partial derivatives, we set each one equal to zero, resulting in a system of equations that needed solving:

• \( 2x - \lambda y = 0 \)
• \( 2y - \lambda x = 0 \)
• \( 1 - xy = 0 \)

These equations represent the conditions where the function is optimized given the constraint. Solving this system ensures that all parts of our original function and constraint are met. With these equations, we were able to deduce possible solutions for \( x \) and \( y \) that minimize our objective function.