Problem 6 Find the local maximum and minim... [FREE SOLUTION]

Chapter 14: Problem 6

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x^{3} y+12 x^{2}-8 y$$

Short Answer

Expert verified

The function has a saddle point at (2, -4).

Step by step solution

Find First-Order Partial Derivatives

To locate critical points, we need the first-order partial derivatives of the function. Let's compute the partial derivative with respect to $x$:\[ f_x(x, y) = \frac{\partial}{\partial x}(x^3y + 12x^2 - 8y) = 3x^2y + 24x \]Similarly, compute the partial derivative with respect to $y$:\[ f_y(x, y) = \frac{\partial}{\partial y}(x^3y + 12x^2 - 8y) = x^3 - 8 \]

Set Partial Derivatives to Zero

Critical points occur where both partial derivatives are zero. Set $f_x(x, y)$ and $f_y(x, y)$ to zero:\[ 3x^2y + 24x = 0 \] \[ x^3 - 8 = 0 \] From $x^3 - 8 = 0$, solve for $x$:\[ x^3 = 8 \Rightarrow x = 2 \]

Solve for Corresponding y Values

Substitute $x = 2$ into the equation $3x^2y + 24x = 0$:\[ 3(2)^2y + 24(2) = 0 \] \[ 12y + 48 = 0 \] \[ 12y = -48 \] \[ y = -4 \]So the critical point is $(2, -4)$.

Determine Nature of Critical Point

Use second-order partial derivatives to analyze the nature of the critical point. First, compute them:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(3x^2y + 24x) = 6xy + 24 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(x^3 - 8) = 0 \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(3x^2y + 24x) = 3x^2 \]Calculate the determinant of the Hessian matrix at the critical point $(x, y) = (2, -4)$:\[ D = f_{xx} f_{yy} - (f_{xy})^2 = (6(2)(-4) + 24)(0) - (3(2)^2)^2 = - (12)^2 = -144 \]Since $D < 0$, the critical point $(2, -4)$ is a saddle point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives

Partial derivatives are fundamental concepts in multivariable calculus. They measure how a function changes as only one variable is altered, keeping the others constant. This is crucial for understanding functions with more than one variable, like our given function.

For a function of two variables, like $f(x, y)$, the partial derivative with respect to $x$ is denoted $f_x(x, y)$. It tells us how $f$ changes as $x$ changes.
Similarly, the partial derivative with respect to $y$, denoted $f_y(x, y)$, shows how $f$ changes as $y$ changes.

In the exercise, we calculated $f_x(x, y) = 3x^2y + 24x$ and $f_y(x, y) = x^3 - 8$. These derivatives are used to locate points where the function does not change instantaneously, known as critical points.

Hessian Matrix

The Hessian matrix is a square matrix composed of second-order partial derivatives of a function. It aids in classifying the nature of critical points for functions of multiple variables.

The elements of the Hessian matrix include $f_{xx}$, $f_{yy}$, and $f_{xy}$, where each represents a second-order partial derivative.
In the exercise, $f_{xx} = 6xy + 24$, $f_{yy} = 0$, and $f_{xy} = 3x^2$.

Once calculated, the Hessian matrix is used to find the determinant $D$, which indicates the type of critical point.
This determinant helps determine whether we have a minimum, maximum, or saddle point.

Saddle Point

A saddle point is a type of critical point in a function where the surrounding surface curves upwards in one direction and downwards in another, like a mountain pass.

It is neither a local minimum nor a maximum.
The determinant of the Hessian matrix, $D$, is used to identify a saddle point.

In our exercise, the determinant $D = -144$ was less than zero, indicating a saddle point at $(2, -4)$.
Saddle points are characterized by their unique topography, unlike peaks or troughs in the graph of a function.

Second-Order Derivatives

Second-order derivatives provide a deeper understanding of the behavior of a function鈥檚 graph.

For multivariable functions, they are essential in forming the Hessian matrix.
Second-order derivatives include $f_{xx}$, $f_{yy}$, and mixed derivatives like $f_{xy}$.

These derivatives measure the rate of change of the first derivatives, offering insight into the function's concavity.
In our solution:

$f_{xx} = 6xy + 24$ mixes $x$ and $y$, indicating interaction between variables.
$f_{yy} = 0$ shows no concavity changes in the $y$ direction alone.
$f_{xy} = 3x^2$ describes how the slope in the $x$ direction changes as $y$ changes.

These insights help us understand the local behavior of the function at critical points.

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91影视

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x^{3} y+12 x^{2}-8 y$$

Short Answer

Step by step solution

Find First-Order Partial Derivatives

Set Partial Derivatives to Zero

Solve for Corresponding y Values

Determine Nature of Critical Point

Key Concepts

Partial Derivatives

Hessian Matrix

Saddle Point

Second-Order Derivatives

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