Problem 321 - Calculus 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. $$ f(x, y)=2 x y+3 x+4 y $$

Step-by-Step Solution

Answer

The critical point \((-2, -1.5)\) is a saddle point.

Step 1: Find the First Partial Derivatives

To use the second derivative test, we first find the partial derivatives of the function \( f(x, y) \). \[ f_x = \frac{\partial}{\partial x}(2xy + 3x + 4y) = 2y + 3 \]\[ f_y = \frac{\partial}{\partial y}(2xy + 3x + 4y) = 2x + 4 \]

Step 2: Set the First Partial Derivatives to Zero

Identify the critical points by setting the partial derivatives to zero:\[ 2y + 3 = 0 \]\[ 2x + 4 = 0 \]Solving these equations, we get:\( y = -\frac{3}{2} \) and \( x = -2 \). Thus, the critical point is \((-2, -1.5)\).

Step 3: Find the Second Partial Derivatives

Calculate the second partial derivatives needed for the second derivative test:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(2xy + 3x + 4y) = 0 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(2xy + 3x + 4y) = 0 \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(2xy + 3x + 4y) = 2 \]

Step 4: Second Derivative Test Calculation

Use the second derivative test formula, which involves the determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 \):\[ D = (0)(0) - (2)^2 = -4 \]

Step 5: Determine the Nature of the Critical Point

Since \( D < 0 \), the point \((-2, -1.5)\) is a saddle point according to the second derivative test.

Key Concepts

Partial Derivatives

Partial derivatives are like regular derivatives but for functions of multiple variables. They help us understand how the function changes when we change just one variable, keeping the others constant.
For the function given in the exercise, we calculated the partial derivatives to explore how the function changes with respect to each variable, **x** and **y**. This is an essential step in finding the critical points.
Here's how the partial derivatives look like for the given function:

• The partial derivative with respect to **x**, denoted as \( f_x \), is calculated by treating **y** as a constant and differentiating the function \( 2xy + 3x + 4y \) with respect to **x**. This gives us \( f_x = 2y + 3 \).
• The partial derivative with respect to **y**, denoted as \( f_y \), is calculated by treating **x** as a constant and differentiating the function with respect to **y**. This results in \( f_y = 2x + 4 \).

By setting these partial derivatives to zero, we are able to identify possible critical points.

Critical Points

Critical points in a function are points at which the first partial derivatives are zero. This means that the rate of change of the function is zero in all directions at these points, making them potential locations for local maxima, minima, or saddle points.
In the given exercise, we set the partial derivatives to zero to find the critical points:

• From \( f_x = 2y + 3 = 0 \), we find \( y = -\frac{3}{2} \).
• From \( f_y = 2x + 4 = 0 \), we find \( x = -2 \).

Thus, the point \((-2, -1.5)\) is a critical point. However, just finding a critical point is not enough to determine its nature; we must further analyze it using the second derivative test.

Saddle Point

A saddle point is a type of critical point where the function does not have a local maximum or minimum. Instead, it resembles a saddle shape where the surface curves up in one direction and down in another.
In the exercise, after finding the critical point, we apply the second derivative test to determine its nature.
The second derivative test involves calculating a determinant, \( D \), which helps to identify the type of critical point. If \( D < 0 \), it signals a saddle point. Since \( D = -4 \) in this case, the critical point \((-2, -1.5)\) is classified as a saddle point.
Saddle points can sometimes be less intuitive than maxima or minima because they don't represent the highest or lowest points locally. Instead, they're more like a change in direction on the surface of the function.

Second Partial Derivatives

Second partial derivatives provide information on the concavity of the function along each variable direction. For the second derivative test, these derivatives are crucial as they help form the determinant used to assess the nature of critical points.
In the example, we need the second partial derivatives to apply the test:

• Calculate \( f_{xx} \), the second partial derivative with respect to **x**, which is \( 0 \) in this case.
• Calculate \( f_{yy} \), the second partial derivative with respect to **y**, also resulting in \( 0 \).
• Calculate \( f_{xy} \), the mixed partial derivative, which is \( 2 \).

The second derivative test uses these to compute the determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 \).
This determinant helps us confirm whether a critical point is a local maximum, minimum, or saddle point by evaluating how the function behaves around these points.