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6. Find the absolute maximum and minimum values of f(x, y) = e¯ region D= {(x, y) : x² + 4y² ≤ 1}. e-xy in the

6. Find the absolute maximum and minimum values of f(x, y) = e¯ region D= {(x, y) : x² + 4y² ≤ 1}. e-xy in the

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
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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6. Absolute maximum value is \( e^{1/4} \), absolute minimum value is \( e^{-1/4} \).
Transcribed Image Text:6. Absolute maximum value is \( e^{1/4} \), absolute minimum value is \( e^{-1/4} \).
### Problem 6: Absolute Maximum and Minimum Values **Objective:** Find the absolute maximum and minimum values of the function \( f(x, y) = e^{-xy} \) within the region defined by \( D \). **Region Definition:** \[ D = \{(x, y) \mid x^2 + 4y^2 \leq 1\} \] This defines an elliptical region in the \( xy \)-plane. **Function:** \[ f(x, y) = e^{-xy} \] To find the absolute maximum and minimum values of the given function \( f(x, y) \) within the region \( D \), one must evaluate the function over the entire region, including boundaries and interior points. The critical points inside the elliptical region, as well as the points on the boundary \( x^2 + 4y^2 = 1 \), should be considered. **Steps to Solve:** 1. **Identify critical points** inside the region by setting up and solving the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) equal to zero. 2. **Evaluate the function on the boundary** of the region defined by \( x^2 + 4y^2 = 1 \). This can often be done using parameterization or Lagrange multipliers. 3. Compare the values of the function at all identified points to determine the absolute maximum and minimum values. **Region Description:** The region \( D \) is an ellipse centered at the origin with a semi-major axis of 1 along the x-axis and a semi-minor axis of 0.5 along the y-axis. This is due to the equation \( x^2 + 4y^2 = 1 \), which is normalized to fit into the unit circle when plotted. **Application:** This problem highlights the method of finding extreme values of a function constrained within a specific region. It's useful in various fields, including optimization, economics, and engineering.
Transcribed Image Text:### Problem 6: Absolute Maximum and Minimum Values **Objective:** Find the absolute maximum and minimum values of the function \( f(x, y) = e^{-xy} \) within the region defined by \( D \). **Region Definition:** \[ D = \{(x, y) \mid x^2 + 4y^2 \leq 1\} \] This defines an elliptical region in the \( xy \)-plane. **Function:** \[ f(x, y) = e^{-xy} \] To find the absolute maximum and minimum values of the given function \( f(x, y) \) within the region \( D \), one must evaluate the function over the entire region, including boundaries and interior points. The critical points inside the elliptical region, as well as the points on the boundary \( x^2 + 4y^2 = 1 \), should be considered. **Steps to Solve:** 1. **Identify critical points** inside the region by setting up and solving the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) equal to zero. 2. **Evaluate the function on the boundary** of the region defined by \( x^2 + 4y^2 = 1 \). This can often be done using parameterization or Lagrange multipliers. 3. Compare the values of the function at all identified points to determine the absolute maximum and minimum values. **Region Description:** The region \( D \) is an ellipse centered at the origin with a semi-major axis of 1 along the x-axis and a semi-minor axis of 0.5 along the y-axis. This is due to the equation \( x^2 + 4y^2 = 1 \), which is normalized to fit into the unit circle when plotted. **Application:** This problem highlights the method of finding extreme values of a function constrained within a specific region. It's useful in various fields, including optimization, economics, and engineering.
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