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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Question
![**Problem 8: Identifying the Critical Point of a Multivariable Function**
The task is to analyze the function \( f(x, y) = x^2 + 4y^2 - 8x + 16y - 2 \) and determine the location and type of its critical point.
To find the critical point, we'll take partial derivatives of the function with respect to \( x \) and \( y \).
1. **Partial Derivative with respect to \( x \):**
- \( \frac{\partial f}{\partial x} = 2x - 8 \)
2. **Partial Derivative with respect to \( y \):**
- \( \frac{\partial f}{\partial y} = 8y + 16 \)
Set both partial derivatives to zero to find critical points:
- \( 2x - 8 = 0 \)
- Solving for \( x \) gives \( x = 4 \).
- \( 8y + 16 = 0 \)
- Solving for \( y \) gives \( y = -2 \).
Thus, the critical point is \((4, -2)\).
To determine the type of critical point, we consider the Hessian matrix:
\[ H = \begin{bmatrix}
\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y}\\
\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}
\end{bmatrix} \]
Where:
- \( \frac{\partial^2 f}{\partial x^2} = 2 \)
- \( \frac{\partial^2 f}{\partial y^2} = 8 \)
- \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)
The Hessian determinant \( D = (2)(8) - (0)^2 = 16 \) which is positive, and since \( \frac{\partial^2 f}{\partial x^2} = 2 \) is positive, the critical point is a local minimum.
**Conclusion:**
The function \( f(x, y) = x^2 + 4y^2 - 8x + 16y - 2 \) has](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa1e36031-6f93-4211-8c57-892d2fd1b551%2F8294fb70-1b13-4ea1-a51c-0f96a6c53843%2Flolxph_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 8: Identifying the Critical Point of a Multivariable Function**
The task is to analyze the function \( f(x, y) = x^2 + 4y^2 - 8x + 16y - 2 \) and determine the location and type of its critical point.
To find the critical point, we'll take partial derivatives of the function with respect to \( x \) and \( y \).
1. **Partial Derivative with respect to \( x \):**
- \( \frac{\partial f}{\partial x} = 2x - 8 \)
2. **Partial Derivative with respect to \( y \):**
- \( \frac{\partial f}{\partial y} = 8y + 16 \)
Set both partial derivatives to zero to find critical points:
- \( 2x - 8 = 0 \)
- Solving for \( x \) gives \( x = 4 \).
- \( 8y + 16 = 0 \)
- Solving for \( y \) gives \( y = -2 \).
Thus, the critical point is \((4, -2)\).
To determine the type of critical point, we consider the Hessian matrix:
\[ H = \begin{bmatrix}
\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y}\\
\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}
\end{bmatrix} \]
Where:
- \( \frac{\partial^2 f}{\partial x^2} = 2 \)
- \( \frac{\partial^2 f}{\partial y^2} = 8 \)
- \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)
The Hessian determinant \( D = (2)(8) - (0)^2 = 16 \) which is positive, and since \( \frac{\partial^2 f}{\partial x^2} = 2 \) is positive, the critical point is a local minimum.
**Conclusion:**
The function \( f(x, y) = x^2 + 4y^2 - 8x + 16y - 2 \) has
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We have to find critical point and its location
Topic:- Application of derivative.
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