Use Lagrange multipliers z = f(x, y) = x² - xy + y² to find the absolute maximum and minimum values of the function subject to the constraint equation x² + y² = 8.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Finding Absolute Maximum and Minimum Using Lagrange Multipliers

In this problem, we aim to find the absolute maximum and minimum values of a given function using the method of **Lagrange multipliers**.

Given:

#### Function:
\[ z = f(x, y) = x^2 - xy + y^2 \]

#### Constraint Equation:
\[ x^2 + y^2 = 8 \]

### Procedure:

1. **Construct the Lagrangian:**

   The Lagrangian function is formed by combining the original function and the constraint equation. It is defined as:
   \[
   \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda \cdot (g(x, y) - c)
   \]
   where \( g(x, y) - c \) represents the constraint equation \( x^2 + y^2 - 8 = 0 \).

   Hence, the Lagrangian for this problem is:
   \[
   \mathcal{L}(x, y, \lambda) = x^2 - xy + y^2 + \lambda(x^2 + y^2 - 8)
   \]

2. **Compute Partial Derivatives:**

   Find the partial derivatives of \(\mathcal{L}(x, y, \lambda)\) with respect to \(x\), \(y\), and \(\lambda\) and set them to zero:
   \[
   \frac{\partial \mathcal{L}}{\partial x} = 0, \quad \frac{\partial \mathcal{L}}{\partial y} = 0, \quad \frac{\partial \mathcal{L}}{\partial \lambda} = 0
   \]

### Solving the System of Equations:

By solving the resulting system of equations, we'll obtain the critical points \((x, y)\). These points will then be evaluated in the original function \( f(x, y) \) to find the absolute maximum and minimum values subjected to the given constraint.

This process will assist students in understanding how to apply Lagrange multipliers for optimization problems in calculus involving constraints.

### Practical Application:

- Students can use this method to tackle real-world problems where optimization with constraints is necessary, such as in economics, engineering, and operations research.

### Graphical Representation:

**Note:** While the current text does not
Transcribed Image Text:### Finding Absolute Maximum and Minimum Using Lagrange Multipliers In this problem, we aim to find the absolute maximum and minimum values of a given function using the method of **Lagrange multipliers**. Given: #### Function: \[ z = f(x, y) = x^2 - xy + y^2 \] #### Constraint Equation: \[ x^2 + y^2 = 8 \] ### Procedure: 1. **Construct the Lagrangian:** The Lagrangian function is formed by combining the original function and the constraint equation. It is defined as: \[ \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda \cdot (g(x, y) - c) \] where \( g(x, y) - c \) represents the constraint equation \( x^2 + y^2 - 8 = 0 \). Hence, the Lagrangian for this problem is: \[ \mathcal{L}(x, y, \lambda) = x^2 - xy + y^2 + \lambda(x^2 + y^2 - 8) \] 2. **Compute Partial Derivatives:** Find the partial derivatives of \(\mathcal{L}(x, y, \lambda)\) with respect to \(x\), \(y\), and \(\lambda\) and set them to zero: \[ \frac{\partial \mathcal{L}}{\partial x} = 0, \quad \frac{\partial \mathcal{L}}{\partial y} = 0, \quad \frac{\partial \mathcal{L}}{\partial \lambda} = 0 \] ### Solving the System of Equations: By solving the resulting system of equations, we'll obtain the critical points \((x, y)\). These points will then be evaluated in the original function \( f(x, y) \) to find the absolute maximum and minimum values subjected to the given constraint. This process will assist students in understanding how to apply Lagrange multipliers for optimization problems in calculus involving constraints. ### Practical Application: - Students can use this method to tackle real-world problems where optimization with constraints is necessary, such as in economics, engineering, and operations research. ### Graphical Representation: **Note:** While the current text does not
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