Problem 326 - Calculus Volume 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$f(x, y)=7 x^{2} y+9 x y^{2}$$

Step-by-Step Solution

Answer

The point \((2/9, -14/9)\) is a local maximum; \((0, 0)\) is inconclusive.

Step 1: Find the First Partial Derivatives

First, we find the first partial derivatives of the function \(f(x, y) = 7x^2 y + 9xy^2\) with respect to \(x\) and \(y\).The first partial derivative with respect to \(x\) is:\[ f_x = \frac{\partial}{\partial x}(7x^2 y + 9xy^2) = 14xy + 9y^2 \]The first partial derivative with respect to \(y\) is:\[ f_y = \frac{\partial}{\partial y}(7x^2 y + 9xy^2) = 7x^2 + 18xy \]

Step 2: Find the Critical Points

To find the critical points, set the first partial derivatives equal to zero and solve the system of equations.Equation 1: \( 14xy + 9y^2 = 0 \)Equation 2: \( 7x^2 + 18xy = 0 \)From Equation 1, \( y(14x + 9y) = 0 \). So, \( y = 0 \) or \( 14x + 9y = 0 \).From Equation 2, \( x(7x + 18y) = 0 \). So, \( x = 0 \) or \( 7x + 18y = 0 \).Solving these, we find that the critical points are \((0, 0)\) and \((2/9, -14/9)\).

Step 3: Calculate the Second Partial Derivatives

Compute the second partial derivatives to use in the second derivative test:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(7x^2 y + 9xy^2) = 14y \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(7x^2 y + 9xy^2) = 18x \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(7x^2 y + 9xy^2) = 14x + 18y \]

Step 4: Apply the Second Derivative Test

Use the second derivative test with the determinant of the Hessian matrix:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]For critical point \((0, 0)\):\[ f_{xx} = 0, \quad f_{yy} = 0, \quad f_{xy} = 0 \]\[ D = (0)(0) - (0)^2 = 0 \]For critical point \((2/9, -14/9)\):\[ f_{xx} = 14(-14/9) = -28/9, \quad f_{yy} = 18(2/9) = 4, \quad f_{xy} = 14(2/9) + 18(-14/9) = -220/9 \]\[ D = \left(\frac{-28}{9}\right)(4) - \left(\frac{-220}{9}\right)^2 \approx 1.8 \]

Step 5: Determine the Nature of Critical Points

For \((0, 0)\), since \(D = 0\), the test is inconclusive; this point could be a saddle point or the function may not have any local extrema here.For \((2/9, -14/9)\), since \(D = 1.8 > 0\) and \(f_{xx} < 0\), the critical point is a local maximum.

Key Concepts

critical points

Critical points are points on a graph where the first partial derivatives of a function are zero. These points are essential when analyzing a function because they are potential locations for local maxima, minima, or saddle points.
To find critical points, calculate the first partial derivatives of the function with respect to each variable. In our exercise, the function is given by \(f(x, y) = 7x^2 y + 9xy^2\).
The first partial derivatives are:

• \(f_x = 14xy + 9y^2\)
• \(f_y = 7x^2 + 18xy\)

By setting these derivatives to zero, you form a system of equations. Solving these equations simultaneously reveals the critical points: \((0, 0)\) and \((2/9, -14/9)\).
These points are then further analyzed to determine their nature using the second derivative test.

Hessian matrix

The Hessian matrix is a square matrix composed of second-order partial derivatives of a scalar function. It is crucial for applying the second derivative test, as it helps determine the concavity of the function near a critical point.
In our exercise, the Hessian matrix \(H\) is derived from the second partial derivatives:

• \(f_{xx} = 14y\)
• \(f_{yy} = 18x\)
• \(f_{xy} = 14x + 18y\)

The Hessian matrix \(H\) is expressed as:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 14y & 14x + 18y \ 14x + 18y & 18x \end{bmatrix}\]Calculating the determinant of this matrix helps to determine the nature of each critical point. The determinant, denoted as \(D\), is used in conjunction with the values of individual second derivatives to ascertain whether a point is a local maximum, local minimum, or a saddle point.

partial derivatives

Partial derivatives are fundamental tools in multivariable calculus, representing how a function changes as only one variable changes, while others remain fixed. These derivatives help in identifying critical points and analyzing the behavior of multivariable functions.
For a function \(f(x, y)\), the partial derivative with respect to \(x\) is found by differentiating \(f\) while keeping \(y\) constant, and vice versa for \(y\).
In our exercise, we calculated the first partial derivatives as follows:

• First partial derivative with respect to \(x\): \(f_x = 14xy + 9y^2\)
• First partial derivative with respect to \(y\): \(f_y = 7x^2 + 18xy\)

Similarly, the second partial derivatives, \(f_{xx}, f_{yy},\) and \(f_{xy}\), provide deeper insight into the function's behavior near critical points. These derivatives are used in the Hessian matrix to apply the second derivative test, playing a critical role in determining the nature of each critical point.

local maximum

A local maximum is a point where a function reaches a peak, such that the function's value at this point is higher than any nearby points. To identify whether a critical point is a local maximum, the second derivative test is used in multivariable calculus.
The second derivative test involves the determinant of the Hessian matrix \(D\). Specifically, a critical point \((x_0, y_0)\) is a local maximum if \(D > 0\) and the second partial derivative \(f_{xx} < 0\) at that point.
In our example, for the critical point \((2/9, -14/9)\), the determinant \(D\) was positive, indicating a local extremum, and \(f_{xx}\) was negative, confirming that it is a local maximum.
Such analysis helps determine the function's behavior at and around critical points, providing vital insights into its graphical representation and practical applications.