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

Step-by-Step Solution

Answer

The critical point \((-2, -\frac{3}{2})\) is a saddle point.

Step 1: Calculate First Derivatives

First, we need to calculate the first partial derivatives of the function \( f(x, y) = 2xy + 3x + 4y \). This gives us the following: \( f_x = \frac{\partial}{\partial x}(2xy + 3x + 4y) = 2y + 3 \) \( f_y = \frac{\partial}{\partial y}(2xy + 3x + 4y) = 2x + 4 \).

Step 2: Find Critical Points

To find the critical points, we set the first partial derivatives to zero and solve the system of equations: \( 2y + 3 = 0 \) \( 2x + 4 = 0 \). From \( f_x = 2y + 3 = 0 \), we find \( y = -\frac{3}{2} \). From \( f_y = 2x + 4 = 0 \), we find \( x = -2 \). Thus, the critical point is \((-2, -\frac{3}{2})\).

Step 3: Calculate Second Derivatives

Next, calculate the second partial derivatives: \( f_{xx} = \frac{\partial^2}{\partial x^2}(2xy + 3x + 4y) = 0 \) \( f_{yy} = \frac{\partial^2}{\partial y^2}(2xy + 3x + 4y) = 0 \) \( f_{xy} = \frac{\partial^2}{\partial x \partial y}(2xy + 3x + 4y) = 2 \).

Step 4: Apply Second Derivative Test

The second derivative test involves calculating the value of \( D = f_{xx} f_{yy} - (f_{xy})^2 \) at the critical point: \( D = 0 \cdot 0 - (2)^2 = -4 \). Since \( D < 0 \), the test tells us that this point is a saddle point.

Key Concepts

Critical Points

In math, a critical point is where the function's derivative is zero or undefined. For functions of two variables, like the given exercise, we find the critical points by

• calculating the first partial derivatives
• setting them to zero
• solving the system of equations

For the function \( f(x, y) = 2xy + 3x + 4y \), this means finding values for \( x \) and \( y \) where the derivative of \( f \) with respect to both variables equals zero.

Using this method, we found a single critical point at \((-2, -\frac{3}{2})\). This point is crucial as it is where potential maxima, minima, or saddle points occur. Understanding how to find critical points is essential because it is the first step in analyzing the behavior of functions in multivariable calculus.

Partial Derivatives

Partial derivatives help us understand how a multivariable function changes with respect to each variable independently. They are crucial in identifying the slope or rate of change of a function in the direction of any coordinate axis.

For a function \( f(x, y) \), the first partial derivative \( f_x \) is found by differentiating \( f \) with respect to \( x \), treating \( y \) as a constant. Similarly, \( f_y \) is obtained by differentiating with respect to \( y \), treating \( x \) as constant.

• For the function given, we calculated \( f_x = 2y + 3 \) and \( f_y = 2x + 4 \).
• These are used to find where the critical points of the function occur.

Partial derivatives are foundational tools in calculus, enabling us to examine and make inferences about a function’s topography in a multidimensional space.

Saddle Point

A saddle point in multivariable calculus appears to be a minimum in one direction and a maximum in another. It doesn’t provide an absolute extremum.

After identifying critical points through partial derivatives, the second derivative test helps decide the nature of these points. A saddle point has unique characteristics defined by the second derivatives. Specifically, the determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \) aids in determining the existence of such points.

• The point \((-2, -\frac{3}{2})\) in the exercise was identified as a saddle point because \( D < 0 \) (in this case, \( D = -4 \)).
• This negative determinant signifies mixed behavior at that point, confirming the saddle nature.

Recognizing a saddle point is fundamental in understanding complex surfaces' topology, offering insight into a function's varied behavior.

Multivariable Calculus

Multivariable calculus expands calculus concepts to functions of several variables. It’s essential for analyzing complex scenarios and has applications in physics, engineering, economics, and more.

• Each function involves several variables, making the study more nuanced compared to single-variable calculus.
• It delves into concepts like partial derivatives, double integrals, and vector calculus, all working together to depict and solve higher-dimensional problems.
• Understanding the behavior of a function such as \( f(x, y) = 2xy + 3x + 4y \) requires the tools provided by multivariable calculus.

By mastering these concepts, students gain the ability to tackle real-world problems involving several changing factors, making multivariable calculus not only a cornerstone of advanced mathematics but also a vital tool in numerous scientific fields.