We appreciate your visit to prove f xy 26x 32x³ 5xy 6y² 12 ST 3x y 170 H3 138. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
The value of [tex]\( |H_{3}| = 3431 \)[/tex] indicates the nature of the critical point [tex]\( (x, y) \)[/tex] where[tex]\( 3x + y = 170 \)[/tex]. Since[tex]\( |H_{3}| > 0 \)[/tex], it means that the critical point (x, y) corresponds to a local minimum for the function f(x, y).
To find the Hessian determinant, we first need to compute the Hessian matrix H of the function f(x, y), and then evaluate its determinant.
1. Compute the Hessian matrix H:
The Hessian matrix \( H \) of a function of two variables is defined as follows:
[tex]\[ H = \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix} \][/tex]
So, for the given function [tex]\( f(x, y) = 26x - 32x^3 + 5xy - 6y^2 + 12 \)[/tex], the second partial derivatives are:
[tex]\[\frac{\partial^2 f}{\partial x^2} = -96x, \quad\frac{\partial^2 f}{\partial y^2} = -12, \quad\frac{\partial^2 f}{\partial x \partial y} = 5, \quad\frac{\partial^2 f}{\partial y \partial x} = 5\][/tex]
2. Evaluate the determinant of H:
The determinant of the Hessian matrix H is denoted as |H|. For the given function, |H| is:
[tex]\[ |H| = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 \][/tex]
Substituting the partial derivatives:
[tex]\[ |H| = (-96x)(-12) - (5)^2 \][/tex]
[tex]\[ |H| = 1152x - 25 \][/tex]
3. Evaluate \( |H| \) at the point where [tex]\( 3x + y = 170 \):[/tex]
Given \3x + y = 170, we can solve for y to express it solely in terms of x: y = 170 - 3x.
Substitute this expression for y into |H|:
[tex]\[ |H| = 1152x - 25 \][/tex]
[tex]\[ |H_{3}| = 1152(3) - 25 = 3456 - 25 = 3431 \][/tex]
So,[tex]\( |H_{3}| = 3431 \).[/tex]
Interpretation:
The value of [tex]\( |H_{3}| = 3431 \)[/tex] indicates the nature of the critical point [tex]\( (x, y) \)[/tex] where[tex]\( 3x + y = 170 \)[/tex]. Since[tex]\( |H_{3}| > 0 \)[/tex], it means that the critical point (x, y) corresponds to a local minimum for the function f(x, y). This means that at the point where 3x + y = 170, the function f(x, y) has a minimum value.
Thanks for taking the time to read prove f xy 26x 32x³ 5xy 6y² 12 ST 3x y 170 H3 138. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
Rewritten by : Barada
