Problem 322 - 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. $$f(x, y)=8 x y(x+y)+7$$

Step-by-Step Solution

Answer

The test is inconclusive at (0,0) as D=0.

Step 1: Find First Derivatives

To apply the second derivative test, the first step is to find the first partial derivatives of the function with respect to both variables, x and y.\[ f_x = \frac{\partial}{\partial x} [8xy(x+y) + 7] = 8y(2x+y)\]\[ f_y = \frac{\partial}{\partial y} [8xy(x+y) + 7] = 8x(x+2y)\]

Step 2: Determine Critical Points

A critical point occurs where both first derivatives are zero. Set \(f_x = 0\) and \(f_y = 0\) and solve simultaneously:1. \(8y(2x+y) = 0\)2. \(8x(x+2y) = 0\)From equation (1), either \(y = 0\) or \(2x+y = 0\).From equation (2), either \(x = 0\) or \(x+2y = 0\).The solutions are: - \((x, y) = (0, 0)\)- Solve \(2x + y = 0\) and \(x + 2y = 0\) by substitution: - From \(2x + y = 0\), \(y = -2x\); - Substitute into \(x + 2(-2x) = 0\): \(x = 0\), \(y = 0\).Thus, the only critical point is \((0, 0)\).

Step 3: Calculate Second Derivatives

Compute the second partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} [8xy(x+y)] = 16y\]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} [8xy(x+y)] = 16x\]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} [8xy(x+y)] = 16x + 8y\]

Step 4: Apply the Second Derivative Test

Use the second derivative test to determine the nature of the critical point \((0, 0)\).Compute the discriminant:\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]At \((0, 0)\): \[ f_{xx}(0,0) = 16(0) = 0, \quad f_{yy}(0,0) = 16(0) = 0, \quad f_{xy}(0,0) = 0\]Thus,\[ D = (0)(0) - (0)^2 = 0 \]

Step 5: Conclude from the Discriminant

Since the discriminant \(D = 0\), the second derivative test is inconclusive. Therefore, at \((0,0)\), the test does not reveal whether this is a maximum, minimum, or saddle point.

Key Concepts

Partial Derivatives

Partial derivatives are a key concept in calculus, especially when dealing with functions of several variables. They help us understand how a function changes as each individual variable changes, while keeping other variables constant. In the context of the exercise, we deal with the function \(f(x, y) = 8xy(x+y) + 7\), which depends on both \(x\) and \(y\).

To find the first partial derivatives, follow these steps:

• For \(f_x\), the derivative with respect to \(x\), treat \(y\) as a constant and differentiate \(8xy(x+y)\). This results in \(f_x = 8y(2x+y)\).
• For \(f_y\), the derivative with respect to \(y\), treat \(x\) as a constant and differentiate \(8xy(x+y)\). This results in \(f_y = 8x(x+2y)\).

Finding these derivatives is essential because they help identify where the function has critical points by setting them equal to zero.

Critical Points

Critical points of a function are the points where its first derivatives are zero, indicating potential maxima, minima, or saddle points. For our function, the critical points are found by setting \(f_x\) and \(f_y\) to zero and solving the resulting equations simultaneously.

For example:

• From \(f_x = 8y(2x+y) = 0\), we find that either \(y = 0\) or \(2x + y = 0\).
• From \(f_y = 8x(x+2y) = 0\), we find that either \(x = 0\) or \(x + 2y = 0\).

Solving these equations, particularly by substitution, we find that the critical point is \((0,0)\). This point is important because it is where the second derivative test is applied to classify the behavior of the function at this location.

Discriminant

The discriminant in the second derivative test helps to categorize critical points as maxima, minima, or saddle points. It is calculated using second partial derivatives. For our function, at the critical point \((0,0)\), we need:

• The second partial derivative \(f_{xx}\) calculated as \(16y\).
• The second partial derivative \(f_{yy}\) calculated as \(16x\).
• The mixed partial derivative \(f_{xy}\) calculated as \(16x + 8y\).

The discriminant \(D\) is given by \(D = f_{xx}f_{yy} - (f_{xy})^2\). At \((0,0)\), both \(f_{xx}\) and \(f_{yy}\) equal zero, and so does \(f_{xy}\), leading to \(D = 0\).

When \(D = 0\), the second derivative test is inconclusive, meaning it doesn’t provide a definitive answer regarding the nature of the critical point. This lack of a conclusion requires us to use other methods or interpretations to understand what happens at \((0,0)\).