**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

Answered: 8. The function f(x, y) = x² + 4y² - 8x…

8. The function f(x, y) = x² + 4y² - 8x + 16y - 2 has one critical point. Determine its location and type.

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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